{"id":626492,"date":"2023-09-12T03:16:58","date_gmt":"2023-09-12T03:16:58","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626492"},"modified":"2023-09-12T03:17:05","modified_gmt":"2023-09-12T03:17:05","slug":"kc-sinha-exercise-19-14-mathematics-solution-class-12-chapter-19-indefinite-integrals","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-19-14-mathematics-solution-class-12-chapter-19-indefinite-integrals\/","title":{"rendered":"KC Sinha: Exercise 19.14- Mathematics Solution Class 12 Chapter 19 Indefinite Integrals"},"content":{"rendered":"\n

<\/span><\/p>\n\n\n\n

Question 1<\/h4>\n\n\n\n

$\\int \\frac{\\cos xdx}{1+9\\sin^2x}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Given : $I=\\int \\frac{\\cos xdx}{1+9\\sin^2 x}$<\/p>\n\n\n\n

$=\\int \\frac{\\cos xdx}{(1)^2+(3\\sin x )^2}$<\/p>\n\n\n\n

3sinx=z then 3cosxdx=dz<\/p>\n\n\n\n

$\\cos xdx=\\frac{1}{3}dz$<\/p>\n\n\n\n

Now $I=\\int \\frac{\\cos xdx}{(1)^2+(3\\sin x)^2}$<\/p>\n\n\n\n

$=\\int \\frac{1}{3(1+z^2)}dx$<\/p>\n\n\n\n

$=\\frac{1}{3}\\int \\frac{1}{1+z^2}dz$<\/p>\n\n\n\n

$=\\frac{1}{3}\\tan^{-1}z+C$<\/p>\n\n\n\n

$=\\frac{1}{3}\\tan^{-1}(3\\sin x)+C$<\/p>\n\n\n\n

Question 2<\/h4>\n\n\n\n

$\\int \\frac{1}{\\sin ^4 x+\\sin^2 c.\\cos ^2 x+\\cos^4 x}dx$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Given: $I=\\int \\frac{1}{\\sin ^4 x+\\sin^2 c.\\cos ^2 x+\\cos^4 x}dx$<\/p>\n\n\n\n

$=\\int \\dfrac{1}{\\frac{\\cos ^4 x}{\\cos ^4 x}\\left(\\sin^4 x+\\sin^2x.\\cos^2 x+\\cos ^4 x\\right)}dx$<\/p>\n\n\n\n

$=\\int \\dfrac{\\sec ^4 x}{\\frac{\\sin^4 x}{\\cos ^4 x}+\\frac{\\sin^2 x.\\cos ^2 x}{\\cos ^2 x.\\cos ^2 x}+\\frac{\\cos ^2 x}{\\cos ^4 x}}dx$<\/p>\n\n\n\n

$=\\iint \\frac{\\sec^2 x.\\sec^2 xdx}{\\tan ^4 x+\\tan^{2}x+1}$<\/p>\n\n\n\n

$=\\int \\frac{(1+\\tan^2 x)\\sec^2 xdx}{\\tan ^4 x+\\tan^2 x+1}$<\/p>\n\n\n\n

Let tanx=z then sec2<\/sup>xdx=dz<\/p>\n\n\n\n

Now $I\\int \\frac{(1+z^2)dz}{z^4+z^2+1}$<\/p>\n\n\n\n

Dividing numerator and denominator by z2<\/sup><\/p>\n\n\n\n

we get $I=\\int \\dfrac{\\frac{1}{z^2}+1}{z^2+1+\\frac{1}{z^2}}dz$<\/p>\n\n\n\n

$=\\int \\dfrac{\\left(1+\\frac{1}{z^2}\\right)}{\\left(z^2+\\frac{1}{z^2}\\right)}dz$<\/p>\n\n\n\n

$\\left[\\therefore \\left(z^2+\\frac{1}{z^2}\\right)=\\left(z-\\frac{1}{z}\\right)^2+2z\\times \\frac{1}{z}\\right]$<\/p>\n\n\n\n

$=\\int \\dfrac{\\left(1+\\frac{1}{z^2}\\right)dz}{\\left(z-\\frac{1}{z}\\right)+2z\\times \\frac{1}{z}+1}$<\/p>\n\n\n\n

$=\\int \\dfrac{\\left(1+\\frac{1}{z^2}\\right)dz}{\\left(z-\\frac{1}{z}\\right)^2+3}$<\/p>\n\n\n\n

$=\\int \\dfrac{\\left(1+\\frac{1}{z^2}\\right)dz}{\\left(z-\\frac{1}{z}\\right)^2+(\\sqrt{3})^2}$<\/p>\n\n\n\n

Let $\\left(z-\\frac{1}{z}\\right)=y$ then $\\left(1+\\frac{1}{z^2}\\right)dz=dy$<\/p>\n\n\n\n

Now $I=\\int \\frac{dy}{z^2+(\\sqrt{3})^2}$<\/p>\n\n\n\n

$=\\frac{1}{\\sqrt{3}}\\tan^{-1}\\frac{y}{\\sqrt{3}}+C$<\/p>\n\n\n\n

$=\\frac{1}{\\sqrt{3}}\\tan^{-1}\\dfrac{\\left(z-\\frac{1}{z}\\right)}{\\sqrt{3}}+C$<\/p>\n\n\n\n

$=\\frac{1}{\\sqrt{3}}\\tan^{-1}\\dfrac{(z^2-1)}{\\sqrt{3}z}+C$<\/p>\n\n\n\n

$I=\\frac{1}{\\sqrt{3}}\\tan^{-1}\\frac{\\tan^{2}x-1}{\\sqrt{3}\\tan x}+C$<\/p>\n\n\n\n

Question 3<\/h4>\n\n\n\n

(i) $\\int \\frac{\\cos 2x.\\sin 2xdx}{\\sqrt{9-\\cos ^4 2x}}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$=\\int \\frac{\\cos 2x .\\sin 2xdx}{\\sqrt{(3)^2-(\\cos ^2 2x)^2}}$<\/p>\n\n\n\n

Let cos2<\/sup>x=z then -2cos2x.sin2x\u00d72dx=dz<\/p>\n\n\n\n

-4cos2x.sin2xdx=dz<\/p>\n\n\n\n

cos2x.sin2xdx$=-\\frac{1}{4}dz$<\/p>\n\n\n\n

Now $I=\\int \\frac{-1dz}{4\\sqrt{(3)^2-(z)^2}}$<\/p>\n\n\n\n

$=\\frac{-1}{4}\\int \\frac{dz}{\\sqrt{(3)^2-z^2}}$<\/p>\n\n\n\n

$=\\frac{-1}{4}\\sin^{-1}\\frac{z}{3}+C$<\/p>\n\n\n\n

$=\\frac{-1}{4}\\sin ^{-1} \\left(\\frac{\\cos^2 2x}{3}\\right)+C$<\/p>\n\n\n\n

(ii) $\\int \\frac{\\cos xdx}{\\sqrt{\\sin^2 x-2\\sin x -3}}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Let sinx=z then cosxdx=dz<\/p>\n\n\n\n

Now $I=\\int \\frac{dz}{\\sqrt{z^2-2z-3}}$<\/p>\n\n\n\n

$=\\int \\frac{dz}{\\sqrt{z^2-2z+1-1-3}}$<\/p>\n\n\n\n

$=\\int \\frac{dz}{\\sqrt{z^2-2z+1-4}}$<\/p>\n\n\n\n

$=\\int \\frac{dz}{\\sqrt{(z-1)^2-(2)^2}}$<\/p>\n\n\n\n

$\\therefore \\int \\frac{dx}{\\sqrt{x^2-a^2}}=\\log|x+\\sqrt{x^2-a^2}|$<\/p>\n\n\n\n

$=\\log|(z-1)+\\sqrt{(z-1)^2-(2)^2}+C|$<\/p>\n\n\n\n

$=log|(\\sin x-1)+\\sqrt{(\\sin x-1)^2-4}|+C$<\/p>\n\n\n\n

$=\\log|(\\sin x-1)+\\sqrt{\\sin^2 x+1-\\sin x-4}|+C$<\/p>\n\n\n\n

$I=\\log|(\\sin x-1)+\\sqrt{\\sin^2 x-2\\sin x-3}|+C$<\/p>\n\n\n\n

(iii)<\/strong> $\\int \\frac{dx}{\\sin^4 x+\\cos ^4 x}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Given : $I=\\int \\frac{dx}{\\sin^4 x+\\cos ^4 x}$<\/p>\n\n\n\n

Dividing numerator and denominator by cos2<\/sup>x<\/p>\n\n\n\n

We get $I=\\int \\dfrac{dx}{\\frac{\\cos ^4 x}{\\cos ^4x }(\\sin^4 x+\\cos ^4 x)}$<\/p>\n\n\n\n

$=\\int \\frac{\\sec^4 xdx}{\\left(\\frac{\\sin^4 x}{\\cos^4 x}+\\frac{\\cos ^4x}{\\cos ^4 x}\\right)}$<\/p>\n\n\n\n

$=\\int \\frac{\\sec^2 x.\\sec^2x dx}{(\\tan^4x+1)}$<\/p>\n\n\n\n

$=\\int \\frac{(1+\\tan^2x)\\sec^2 xdx}{\\tan^4 x+1}$<\/p>\n\n\n\n

Let tanx=z then sec2<\/sup>xdx=dz<\/p>\n\n\n\n

Now $=\\int \\frac{(1+z^2)dz}{z^4+1}$<\/p>\n\n\n\n

Dividing numerator and denominator by z2<\/sup><\/p>\n\n\n\n

We get $I=\\int \\dfrac{\\left(\\frac{1}{z^2}+1\\right)dz}{\\left(z^2+\\frac{1}{z^2}\\right)}$<\/p>\n\n\n\n

$=\\int \\dfrac{\\left(1+\\frac{1}{z^2}\\right)dz}{\\left(z-\\frac{1}{z}\\right)^2+2}$<\/p>\n\n\n\n

Let $ \\left(z-\\frac{1}{z}\\right)=y$ then $\\left(1+\\frac{1}{z^2}\\right)dz=dy$<\/p>\n\n\n\n

Now $I=\\int \\frac{dy}{y^2+2}$<\/p>\n\n\n\n

$=\\int \\frac{dy}{y^2+(\\sqrt{2})^2}$<\/p>\n\n\n\n

$=\\frac{1}{\\sqrt{2}}\\tan^{-1}\\left(\\frac{y}{\\sqrt{2}}\\right)+C$<\/p>\n\n\n\n

$=\\frac{1}{\\sqrt{2}}\\tan^{-1}\\dfrac{\\left(z-\\frac{1}{z}\\right)}{\\sqrt{2}}+C$<\/p>\n\n\n\n

$=\\dfrac{1}{\\sqrt{2}}\\tan^{-1}\\left(\\dfrac{z^2-1}{\\sqrt{2}z}\\right)+C$<\/p>\n\n\n\n

$=\\frac{1}{\\sqrt{2}}\\tan^{-1}\\frac{(\\tan^{-1}x-1)}{\\sqrt{2}\\tan x}+C$<\/p>\n\n\n\n

(iv)<\/strong> $\\int \\frac{x^2+4}{x^4+16}dx$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Given : $I=\\int \\frac{x^2+4}{x^4+16}dx$<\/p>\n\n\n\n

Dividing numerator and denominator by x2<\/sup><\/p>\n\n\n\n

we get<\/p>\n\n\n\n

$I=\\int \\dfrac{1+\\frac{4}{x^2}}{x^2+\\frac{16}{x^2}}dx$<\/p>\n\n\n\n

$=\\int \\dfrac{\\left(1+\\frac{4}{x^2}\\right)dx}{\\left(x-\\frac{4}{x}\\right)^2+8}$<\/p>\n\n\n\n

$=\\int \\dfrac{\\left(1+\\frac{4}{x^2}\\right)dx}{\\left(x-\\frac{4}{x}\\right)^2+\\left(2\\sqrt{2}^2\\right)}$<\/p>\n\n\n\n

Let $\\left(x-\\frac{4}{x}\\right)=z$ then $\\left(1+\\frac{4}{x^2}\\right)dx=dz$<\/p>\n\n\n\n

Now $I=\\int \\frac{dz}{z^2+(2\\sqrt{2})^2}$<\/p>\n\n\n\n

$=\\frac{1}{2\\sqrt{2}}\\tan^{-1}\\left(\\frac{z}{2\\sqrt{2}}\\right)+C$<\/p>\n\n\n\n

$=\\frac{1}{2\\sqrt{2}}\\tan^{-1}\\dfrac{\\left(x-\\frac{4}{x}\\right)}{2\\sqrt{2}}+C$<\/p>\n\n\n\n

$=\\frac{1}{2\\sqrt{2}}\\tan^{-1}\\left(\\frac{x^2-4}{2\\sqrt{2}x}\\right)+C$<\/p>\n\n\n\n

Question 4<\/h4>\n\n\n\n

$\\int \\frac{\\cos xdx}{\\sin x+\\cos x}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Let cosx=A(sinx+cosx)+B$\\frac{d}{dx}$(sinx+cosx)<\/p>\n\n\n\n

=cosx=A(sinx+cosx)+B(cosx-sinx)<\/p>\n\n\n\n

=cosx=Asinx+Acosx+Bcosx-Bsinx<\/p>\n\n\n\n

=cosx=(A-B)sinx+(A+B)cosx<\/p>\n\n\n\n

Equating the coefficient of sinx and cosx we get<\/p>\n\n\n\n

A-B=0<\/p>\n\n\n\n

A=B<\/p>\n\n\n\n

and A+B=1<\/p>\n\n\n\n

\u21d2A+A=1<\/p>\n\n\n\n

\u21d22A=1<\/p>\n\n\n\n

\u21d2$A=\\frac{1}{2}$<\/p>\n\n\n\n

\u21d2$B=\\frac{1}{2}$<\/p>\n\n\n\n

Now $I=\\int \\frac{\\cos xdx}{\\sin x+\\cos x}$<\/p>\n\n\n\n

$=\\int \\frac{A(\\sin x+\\cos x )+B(\\cos x-\\sin x)}{\\sin x+\\cos x}$<\/p>\n\n\n\n

$=A\\int \\frac{\\sin x+\\cos xdx}{\\sin x +\\cos x}+B\\int \\frac{\\cos x-\\sin dx}{\\sin x+\\cos x}$<\/p>\n\n\n\n

$=\\frac{1}{2}\\int dx+\\frac{1}{2}\\int \\frac{dz}{z}$<\/p>\n\n\n\n

$=\\frac{1}{2}x+\\frac{1}{2}\\log|z|+C$<\/p>\n\n\n\n

$=\\frac{x}{2}+\\frac{1}{2}\\log|\\sin x +\\cos x|+C$<\/p>\n\n\n\n

Question 5<\/h4>\n\n\n\n

$\\int \\frac{\\cos xdx}{2\\sin x+3\\cos x}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Let cosx=A(2sinx+3cosx)+B$\\frac{d}{dx}$(2sinx+3cosx)<\/p>\n\n\n\n

=cosx=A(2sinx+3cosx)+B(2cosx-3sinx)<\/p>\n\n\n\n

=cosx=2Asinx+3Acosx+2Bcosx-2Bsinx<\/p>\n\n\n\n

=1.cosx=(2A-3B)sinx+(3A+2B)cosx<\/p>\n\n\n\n

Equating the coefficient of sinx and cosx<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

2A=3B=0 and 3A+2B=1<\/p>\n\n\n\n

$\\begin{array}{l|l}2A=3B&3.\\frac{3}{2}B+2B=1\\\\\\therefore A=\\frac{3}{2}B& \\frac{9B+4B}{2}=1\\\\\\therefore A=\\frac{3}{2}\\times \\frac{2}{13}& 13B^2=2\\\\A=\\frac{3}{13}&B=\\frac{2}{13}\\end{array}$<\/p>\n\n\n\n

Now $I=\\int \\frac{\\cos xdx}{2\\sin x +3\\cos x}$<\/p>\n\n\n\n

$=\\int \\frac{A(2\\sin x+3\\cos x)+B(2\\cos x-3\\sin x)}{2\\sin x+3\\cos x}$<\/p>\n\n\n\n

$=A\\frac{2\\sin x+3 \\cos xdx}{2\\sin x+3\\cos x}+B\\int {2\\cos x-3\\sin x dx}{2\\sin x+3\\cos x}$<\/p>\n\n\n\n

$=\\frac{3}{13}\\int dx+\\frac{2}{13}\\int \\frac{dz}{z}$<\/p>\n\n\n\n

$I=\\frac{3}{13}x+\\frac{2}{13}\\log|z|+C$<\/p>\n\n\n\n

$=\\frac{3}{13}x+\\frac{2}{13}\\log|2\\sin x+3\\cos x|+C$<\/p>\n\n\n\n

Question 6<\/h4>\n\n\n\n

$\\int \\frac{dx}{1+\\tan x}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Given : $I=\\int \\frac{dx}{1+\\tan x}$<\/p>\n\n\n\n

$=\\int \\frac{dx}{1+\\frac{\\sin x }{\\cos x}}$<\/p>\n\n\n\n

$=\\int \\frac{\\cos xdx}{\\cos x+\\sin x}$<\/p>\n\n\n\n

$I=\\frac{x}{2}+\\frac{1}{2}\\log|\\cos x+\\sin x|+C$<\/p>\n\n\n\n

Question 7<\/h4>\n\n\n\n

$\\int \\frac{dx}{1+\\cot x}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Given : $I=\\int \\frac{dx}{1+\\cot x}$<\/p>\n\n\n\n

$=\\int \\frac{dx}{1+\\frac{\\cos x}{\\sin x}}$<\/p>\n\n\n\n

$=\\int \\frac{\\sin xdx}{\\sin x +\\cos x}$<\/p>\n\n\n\n

Let sinx=A(sinx+cosx)+B$\\frac{d}{dx}$(sinx+cosx)<\/p>\n\n\n\n

=sinx=A(sinx+cosx)+B(cosx-sinx)<\/p>\n\n\n\n

=sinx=Asinx+Acosx+Bcosx-Bsinx<\/p>\n\n\n\n

=sinx=(A-B)sinx+(A+B)cosx<\/p>\n\n\n\n

Equating the coefficient of sinx and cosx <\/p>\n\n\n\n

we get $\\begin{array}{l|l}A-B=1&A+B=0\\\\-B-b+1&\\therefore A=-B\\\\-2B=1&A=-\\left(-\\frac{1}{2}\\right)=\\frac{1}{2}\\\\B=\\frac{-1}{2}&\\end{array}$<\/p>\n\n\n\n

Now $I=\\int \\frac{\\sin xdx}{\\sin x+cos x}$<\/p>\n\n\n\n

$=\\int \\frac{A(\\sin x+\\cos x)+B(\\cos x-\\sin x)dx}{\\sin x+\\cos x}$<\/p>\n\n\n\n

$=A\\int \\frac{\\sin x+\\cos xdx}{\\sin x+\\cos x}+B\\int \\frac{\\cos x-\\sin xdx}{\\sin x+\\cos x}$<\/p>\n\n\n\n

$=A\\int dx+B\\frac{dz}{z}$<\/p>\n\n\n\n

$I=\\frac{1}{2}x-\\frac{1}{2}\\log|z|+C$<\/p>\n\n\n\n

$I=\\frac{x}{2}-\\frac{1}{2}\\log|\\sin x+\\cos x|+C$<\/p>\n\n\n\n

Question 8<\/h4>\n\n\n\n

(i)<\/strong> $\\int \\frac{\\sin x+2\\cos xdx}{2\\sin x+\\cos x}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Let sinx+2cosx=A(2sinx+cosx)+B$\\frac{d}{dx}$(2sinx+cosx)<\/p>\n\n\n\n

=sinx+2cosx=A(2sinx+cosx)+B(2cosx-sinx)<\/p>\n\n\n\n

=sinx+2cosx=2Asinx+Acosx+2Bcosx-Bsinx<\/p>\n\n\n\n

=sinx+2cosx=(2A-B)sinx+(A+2B)cosx<\/p>\n\n\n\n

Equating the coefficient of sinx and cosx<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

$\\begin{array}{l|l}2A-B=1&A+2B=2\\\\2A=1+B&\\frac{1+B}{2}+2B=2\\\\A=\\frac{1+B}{2}&\\frac{1+B+4B}{2}=2\\\\ \\therefore A=\\frac{1+\\frac{3}{5}}{2}=\\frac{8}{10}&5B=4-1\\\\ \\therefore A=\\frac{4}{5}& B=\\frac{3}{5}\\end{array}$<\/p>\n\n\n\n

Now $I=\\int \\frac{\\sin x+2\\cos xdx}{2\\sin x+\\cos x}$<\/p>\n\n\n\n

$=\\int \\frac{A(2\\sin x+\\cos x)+B(2\\cos x-\\sin x)}{2\\sin x+\\cos x}$<\/p>\n\n\n\n

$=A\\int \\frac{2\\sin x+\\cos xdx}{2\\sin x+\\cos x}+B\\int \\frac{2\\cos x-\\sin xdx}{2\\sin x+\\cos x }dx$<\/p>\n\n\n\n

$=A\\int dx+B\\int \\frac{dz}{z}$<\/p>\n\n\n\n

=Ax+Blogz+C<\/p>\n\n\n\n

$I=\\frac{4}{5}x+\\frac{3}{5}\\log|2\\sin x+\\cos x|+C$<\/p>\n\n\n\n

(ii)<\/strong> $\\int \\frac{\\sin xdx}{3\\sin x+5\\cos x}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Let sinx=A(3sinx+5cosx)<\/p>\n\n\n\n

Let sinx=A(3sinx+5cosx)+B$\\frac{d}{dx}$(3sinx+5cosx)<\/p>\n\n\n\n

=sinx=A(3sinx+5cosx)+B(3cosx-5sinx)<\/p>\n\n\n\n

=sinx=3Asinx+5Acosx+3Bcosx-5Bsinx<\/p>\n\n\n\n

=sinx=(3A-5B)sinx+(5A+3B)cosx<\/p>\n\n\n\n

Equating the coefficient of sinx and cosx  <\/p>\n\n\n\n

We get<\/p>\n\n\n\n

$\\begin{array}{l|l}3A-5B=1& 5A+3B=0\\\\\\therefore 3\\left(-\\frac{3}{5}B\\right)-5B=1&5A-3B\\\\-\\frac{9B}{5}-5B=1&A=\\frac{-3}{5}B\\\\\\frac{-9B-25B}{5}=1& \\therefore A=\\frac{-3}{5}\\times \\frac{-5}{34}\\\\-34B=5&A=\\frac{3}{34}\\\\\\therefore B=-\\frac{5}{34}&\\end{array}$<\/p>\n\n\n\n

Now $I=\\int \\frac{\\sin xdx}{3\\sin x+5\\cos x}$<\/p>\n\n\n\n

$=\\int \\frac{A(3\\sin x+5\\cos x)+B(3\\cos x-5\\sin x)}{3\\sin x+5\\cos x}$<\/p>\n\n\n\n

$=A\\int \\frac{3\\sin x+5\\cos x dx}{3\\sin x+5\\cos x}+B\\int \\frac{3\\cos x-5\\cos xdx}{3\\sin x+5\\cos x}$<\/p>\n\n\n\n

$=A\\int dx+B\\int \\frac{dz}{z}$<\/p>\n\n\n\n

$=\\frac{3}{34}x-\\frac{5}{34}\\log|z|+C$<\/p>\n\n\n\n

$=\\frac{3}{34}x-\\frac{5}{34}\\log|3\\sin x+5\\cos x|+C$<\/p>\n\n\n\n

Question 9<\/h4>\n\n\n\n

$\\int \\frac{dx}{1+\\cos^2 x}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Given : $I=\\int \\frac{dx}{1+\\cos^2 x}$<\/p>\n\n\n\n

$=\\int \\dfrac{dx}{\\frac{\\cos^2 x}{\\cos^2 x}(1+\\cos^2 x)}$<\/p>\n\n\n\n

$=\\int \\dfrac{\\sec^2 xdx}{\\frac{1}{\\cos^2 x}+\\frac{\\cos^2 x}{\\cos^2 x}}$<\/p>\n\n\n\n

$=\\int \\frac{\\sec^2 xdx}{\\sec^2 x+1}$<\/p>\n\n\n\n

$=\\int \\frac{\\sec^2 xdx}{1+\\tan^2 x+1}$<\/p>\n\n\n\n

$=\\int \\frac{\\sec^2 xdx}{2+\\tan^2 x}$<\/p>\n\n\n\n

$=\\int \\frac{\\sec^2 xdx}{(\\sqrt{2})^2+(\\tan x)}$<\/p>\n\n\n\n

Let tanx=z then sec2<\/sup>xdx=dz<\/p>\n\n\n\n

Now $I=\\int \\frac{\\sec^2 xdx}{(\\sqrt{2})^2+(\\tan x)^2}$<\/p>\n\n\n\n

$=\\int \\frac{dz}{(\\sqrt{2})^2+z^2}$<\/p>\n\n\n\n

$=\\frac{1}{\\sqrt{2}}\\tan^{-1}+\\frac{z}{\\sqrt{2}}+C$<\/p>\n\n\n\n

$=\\frac{1}{\\sqrt{2}}\\tan^{-1}\\frac{(\\tan x)}{\\sqrt{2}}+C$<\/p>\n\n\n\n

Question 10<\/h4>\n\n\n\n

$\\int \\frac{dx}{4\\sin^2x+9\\cos^2 x}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Given : $I=\\int \\frac{dx}{4\\sin^{2}x+9\\cos^2 x}$<\/p>\n\n\n\n

$=\\dfrac{dx}{\\frac{\\cos^2 x}{\\cos^2 x}(4\\sin^2 x+9\\cos^2 x)}$<\/p>\n\n\n\n

$=\\int \\dfrac{\\sec^2 xdx}{\\left(\\frac{4\\sin^2 x}{\\cos^2 x}+\\frac{9\\cos^2 x}{\\cos^2 x}\\right)}$<\/p>\n\n\n\n

$=\\int \\frac{\\sec^2 xdx}{4\\tan^{2}x+9}$<\/p>\n\n\n\n

Let tanx=z then sec2<\/sup>xdx=dz<\/p>\n\n\n\n

Now $I=\\int \\frac{dz}{4z^2+9}$<\/p>\n\n\n\n

$=\\int \\frac{dz}{(2z)^2+(3)^2}$<\/p>\n\n\n\n

$\\left[\\therefore \\int \\frac{dx}{x^2+a^2}=\\frac{1}{a}\\tan^{-1}\\frac{x}{a}\\right]$<\/p>\n\n\n\n

$=\\frac{1}{3}\\tan^{-1}\\left(\\frac{2z}{3}\\right)+C$<\/p>\n\n\n\n

$=\\frac{1}{3\\times 2}\\tan^{-1}\\left(\\frac{2}{3}\\tan x\\right)+C$<\/p>\n\n\n\n

$I=\\frac{1}{6}\\tan^{-1}\\left(\\frac{2}{3}\\tan x\\right)+C$<\/p>\n\n\n\n

Question 11<\/h4>\n\n\n\n

$\\int \\frac{dx}{(a\\sin x+b\\cos x)}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Given : $I=\\int \\frac{dx}{(a\\sin x+b\\cos x)}$<\/p>\n\n\n\n

$=\\int \\frac{dx}{a^2\\sin^2 x+b^2 \\cos ^2 x+2ab\\sin x. \\cos x}$<\/p>\n\n\n\n

$=\\int \\dfrac{dx}{\\frac{\\cos ^2 x}{\\cos ^2 x}\\left(a^2 \\sin ^2 x+b^2 \\cos ^2 x+2ab \\sin x.\\cos x\\right)}$<\/p>\n\n\n\n

$=\\int \\dfrac{\\sec ^2 xdx}{\\frac{a^2 \\sin ^2 x}{\\cos ^2 x}+\\frac{b^2 \\cos ^2 x}{\\cos ^2 x}+\\frac{2ab\\sin x.\\cos x}{\\cos ^2 x}}$<\/p>\n\n\n\n

$=\\int \\frac{\\sec^2 xdx}{a^2 \\tan^2 x+b^2 +2ab\\tan x}$<\/p>\n\n\n\n

Let tanx=z then sec2<\/sup>xdx=dz<\/p>\n\n\n\n

Now $I=\\int \\frac{dz}{a^2z^2+b^2+2abz}$<\/p>\n\n\n\n

$=\\int \\frac{dz}{(az+b)^2}$<\/p>\n\n\n\n

$=\\int (az+b)^{-2}dz$<\/p>\n\n\n\n

$=\\frac{(az+b)^{-2+1}}{(-2+1)a}+C$<\/p>\n\n\n\n

$=\\frac{(az+b)^{-1}}{-1\\times a}+C$<\/p>\n\n\n\n

$=\\frac{-1}{a(az+b)}+C$<\/p>\n\n\n\n

$=\\frac{-1}{a(a\\tan x+b)}+C$<\/p>\n\n\n\n

Question 12<\/h4>\n\n\n\n

$\\int \\frac{dx}{5+4\\sin x}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Given : $I=\\int \\frac{dx}{5+4\\sin x}$<\/p>\n\n\n\n

$\\left[\\therefore \\sin x=\\dfrac{2\\tan \\frac{x}{2}}{1+\\tan^2\\frac{x}{2}}\\right]$<\/p>\n\n\n\n

$=\\int 5+4\\left(\\dfrac{2\\tan \\frac{x}{2}}{1+\\tan ^2 \\frac{x}{2}}\\right)$<\/p>\n\n\n\n

$=\\int \\frac{1+\\tan^{2}\\frac{x}{2}dx}{5+5\\tan^2 \\frac{x}{2}+8\\tan \\frac{x}{2}}$<\/p>\n\n\n\n

$=\\int \\frac{1+\\tan^2 \\frac{x}{2}dx}{5+5\\tan^2 \\frac{x}{2}+8\\tan \\frac{x}{2}}$<\/p>\n\n\n\n

$=\\int \\frac{\\sec^2 \\frac{x}{2}}{5\\tan^2 \\frac{x}{2}+8\\tan \\frac{x}{2}+5}$<\/p>\n\n\n\n

Let $\\tan \\frac{x}{2}=z$ then $\\frac{1}{2}\\sec^2 \\frac{x}{2}=dz$<\/p>\n\n\n\n

$\\sec^2 \\frac{x}{2}dx=2dz$<\/p>\n\n\n\n

Now $I=\\int \\frac{2dz}{5z^2+8z+5}$<\/p>\n\n\n\n

$=2\\int \\frac{dz}{5\\left(z^2+\\frac{8}{5}z+1\\right)}$<\/p>\n\n\n\n

$=\\frac{2}{5}\\int \\dfrac{dz}{z^2+2.z.\\frac{4}{5}+\\frac{16}{25}-\\frac{16}{25}+1}$<\/p>\n\n\n\n

$=\\frac{2}{5}\\int \\dfrac{dz}{\\left(z+\\frac{4}{5}\\right)^2+\\frac{9}{25}}$<\/p>\n\n\n\n

$=\\frac{2}{5}\\int \\frac{1dz}{\\left(z+\\frac{4}{5}\\right)^2+\\left(\\frac{3}{5}\\right)^2}$<\/p>\n\n\n\n

$=\\frac{2}{5}\\times \\dfrac{1}{\\frac{3}{5}} \\tan ^{-1} \\dfrac{\\left(z+\\frac{4}{5}\\right)}{\\frac{3}{5}}+C$<\/p>\n\n\n\n

$=\\frac{2}{5}\\times \\frac{5}{3}\\tan^{-1} \\dfrac{\\left(5z+4\\right)}{\\frac{3}{5}\\times 5}+C$<\/p>\n\n\n\n

$=\\frac{2}{5}\\tan ^{-1}\\left(\\frac{5z+4}{3}\\right)+C$<\/p>\n\n\n\n

$=\\frac{2}{3}\\tan^{-1}\\left(\\frac{5\\tan \\frac{x}{2}+4}{3}\\right)+C$<\/p>\n\n\n\n

Question 13<\/h4>\n\n\n\n

$\\int \\frac{dx}{4+5\\cos x}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Given : $I=\\int \\frac{dx}{4+5\\cos x}$<\/p>\n\n\n\n

$=\\int \\dfrac{dz}{4+5\\left(\\dfrac{1-\\tan ^2\\frac{x}{2}}{1+\\tan ^2 \\frac{x}{2}}\\right)}$<\/p>\n\n\n\n

$=\\int \\frac{dx \\sec^2 \\frac{x}{2}}{4+4\\tan ^2 \\frac{x}{2}+5-5\\tan ^2 \\frac{x}{2}}$<\/p>\n\n\n\n

$=\\int \\frac{\\sec^2 \\frac{x}{2}dx}{9-\\tan ^2 \\frac{x}{2}}$<\/p>\n\n\n\n

Let $\\tan \\frac{x}{2}=z$ then $\\frac{1}{2}\\sec^2 \\frac{x}{2}dx=dz$ <\/p>\n\n\n\n

$\\therefore \\sec^2 \\frac{x}{2}dx=2dz$<\/p>\n\n\n\n

Now $I=\\int \\frac{2dz}{9-z^2}$<\/p>\n\n\n\n

$=\\int \\frac{2dz}{(3)^2-z^2}$<\/p>\n\n\n\n

$=2\\int \\frac{dz}{(3)^2-(z)^2}$<\/p>\n\n\n\n

$\\left[\\int \\frac{dx}{a^2-x^2}=\\frac{1}{2a}\\log\\left|\\frac{a+x}{a-x}\\right|\\right]$<\/p>\n\n\n\n

$=2\\times \\frac{1}{2\\times 3} \\log\\left|\\frac{3+z}{3-z}\\right|+C$<\/p>\n\n\n\n

$=\\frac{1}{3}\\log\\left|\\dfrac{3+\\tan\\frac{x}{2}}{3-\\tan \\frac{x}{2}}\\right|+C$<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 12 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 1 $\\int \\frac{\\cos xdx}{1+9\\sin^2x}$ Sol : Given : $I=\\int \\frac{\\cos xdx}{1+9\\sin^2 x}$ $=\\int \\frac{\\cos xdx}{(1)^2+(3\\sin x )^2}$ 3sinx=z then 3cosxdx=dz $\\cos xdx=\\frac{1}{3}dz$ Now $I=\\int \\frac{\\cos xdx}{(1)^2+(3\\sin x)^2}$ $=\\int \\frac{1}{3(1+z^2)}dx$ $=\\frac{1}{3}\\int \\frac{1}{1+z^2}dz$ $=\\frac{1}{3}\\tan^{-1}z+C$ $=\\frac{1}{3}\\tan^{-1}(3\\sin x)+C$ Question 2 $\\int \\frac{1}{\\sin ^4 x+\\sin^2 c.\\cos ^2 x+\\cos^4 x}dx$ Sol : Given: $I=\\int \\frac{1}{\\sin ^4 x+\\sin^2 c.\\cos ^2 x+\\cos^4 […]<\/p>\n","protected":false},"author":302,"featured_media":626470,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[25],"tags":[],"boards":[],"class_list":["post-626492","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-12","entry"],"yoast_head":"\nKC Sinha: Exercise 19.14- Mathematics Solution Class 12 Chapter 19 Indefinite Integrals - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1 $int frac{cos xdx}{1+9sin^2x}$ Sol : Given : $I=int frac{cos xdx}{1+9sin^2 x}$ $=int frac{cos xdx}{(1)^2+(3sin x )^2}$ 3sinx=z then\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-19-14-mathematics-solution-class-12-chapter-19-indefinite-integrals\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 19.14- Mathematics Solution Class 12 Chapter 19 Indefinite Integrals\" \/>\n<meta property=\"og:description\" content=\"Question 1 $int frac{cos xdx}{1+9sin^2x}$ Sol : Given : $I=int frac{cos xdx}{1+9sin^2 x}$ $=int frac{cos xdx}{(1)^2+(3sin x )^2}$ 3sinx=z then\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-19-14-mathematics-solution-class-12-chapter-19-indefinite-integrals\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" 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class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-19-14-mathematics-solution-class-12-chapter-19-indefinite-integrals\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-19-14-mathematics-solution-class-12-chapter-19-indefinite-integrals\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"KC Sinha: Exercise 19.14- Mathematics Solution Class 12 Chapter 19 Indefinite 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