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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the integral using substitution $\int\limits_0^\frac{\large\pi}{2}\sqrt{\sin\phi}\cos^5\phi\;d\phi$

$\begin{array}{1 1}\large\frac{64}{231} \\ \large\frac{32}{231} \\\large\frac{16}{231} \\ \large\frac{96}{231} \end{array} $

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
$\int\limits_0^\frac{\large\pi}{2}\sqrt{\sin\phi}\cos^5\phi\;d\phi$
$\Rightarrow \int\limits_0^{\Large\frac{\pi}{2}}\sqrt{\sin\phi}\cos^5\phi d\phi$
$\Rightarrow \int\limits_0^{\Large\frac{\pi}{2}}\sqrt{\sin\phi}(1-\sin^2\phi)^2\cos\phi d\phi$
$\cos^2\theta+\sin^2\theta=1$
Step 2:
Substitute $\sin\phi =t$
$\cos\phi =\large\frac{dt}{d\phi}$
Differentiating with respect to $\phi$
$d\phi=\large\frac{dt}{\cos\phi}$
Step 3:
For limit when $\phi=0\Rightarrow t=1$
When $\phi=\large\frac{\pi}{2}$
$t=1$
$I=\int \limits_0^1\sqrt t(1-t^2)^2dt$
$\;\;=\int \limits_0^1\sqrt t(t^4+1-2t^2)dt$
$\;\;\;=\int \limits_0^1(t^{\Large\frac{9}{2}}+t^{\Large\frac{1}{2}}-2t^{\Large\frac{5}{2}})$$dt$
$\;\;\;=[\large\frac{2}{11}+\frac{2}{3}+\frac{4}{7}]-0$
$\;\;\;=\large\frac{42+154-132}{11\times 3\times 7}=\frac{64}{231}$
answered Sep 10, 2013 by sreemathi.v
 
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