Browse Questions

# Evaluate the integral using substitution $\int\limits_0^\frac{\large\pi}{2}\sqrt{\sin\phi}\cos^5\phi\;d\phi$

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

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$
$\;\;\;=[\large\frac{2}{11}+\frac{2}{3}+\frac{4}{7}]-0$
$\;\;\;=\large\frac{42+154-132}{11\times 3\times 7}=\frac{64}{231}$