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If  \int \frac{\cos \: x\: dx}{\sin ^{3}x(1+\sin ^{6}x)^{2/3}}= f(x)(1+ \sin ^{6}x)^{1/\lambda }+c where c is a constant of integration, then \lambda f\left ( \frac{\pi }{3} \right ) is equal to :
Option: 1 -\frac{9}{8}
Option: 2 \frac{9}{8}
Option: 3 2
Option: 4 -2
 

Answers (1)

best_answer

 

 

Integration Using Substitution -

The method of substitution is one of the basic methods for calculating indefinite integrals. 

Substitution - change of variable

\\\mathrm{To\;solve\;the\;integrate\;of\;the\;form}\\\\\mathrm{I=\int f\left ( g(x) \right )\cdot g'(x)\;dx,\;}\\\\\mathrm{\;where\;g(x)\;is\;continuously\;differentiable\;function.}\\\mathrm{put\;\;g(x)=t,\;\;g'(x)\;dx=dt}\\\mathrm{After\;substitution,\;we\;get\;\;\int f(t)\;dt.}\\\text{Evalute this integration and substitute back the value of }t.

 

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\\\sin x=t \quad\Rightarrow \cos x d x=d t\\\\\int \frac{d t}{t^{2}\left(1+t^{6}\right)^{2 / 3}}=\int \frac{d t}{t^{3} t^{4}\left(\frac{1}{t^6}+1\right)^{2/3}}\\\\u=\frac{1}{t^{6}}+1 \Rightarrow d u=-\frac{6}{t^{7}} d t\\\\\frac{d u}{-6}=\frac{d t}{t^{7}}\\

\\=\int\frac{du}{-6u^{2/3}}=-\frac{1}{2 }u^{1/3}+C\\\\=-\frac{1}{2}\left ( \frac{1}{t^6}+1 \right )^{1/3}+C\\\\=-\frac{1}{2}\left ( \frac{(1+\sin^6x)^{1/3}}{\sin^2x} \right )\\\\f(x)=-\frac{1}{2}\frac{1}{\sin^2x}\;\;\;\lambda=3\\\\\lambda f(\pi/3)=-2

Correct option (4)

Posted by

Kuldeep Maurya

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