Question

Evaluate the following integral : $\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{3}}xdx}$

Answer 1

Hint: Convert \[{{\sin }^{2}}x=1-{{\cos }^{2}}x\]${{\sin }^{3}}x=\sin x\left( 1-{{\cos }^{2}}x \right)$and solve as two separate integrals and later add it.

Complete step-by-step answer:

We are given the following definite integral to evaluate,

$\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{3}}xdx}$

Now, let’s consider the integral as ‘I’.

So,

I=$\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{3}}xdx}$

Now, we will transform I as,

We know that, \[{{\sin }^{2}}x=1-{{\cos }^{2}}x\]

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x\left( 1-{{\cos }^{2}}x \right)}dx$

\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{(\sin x-\sin x{{\cos }^{2}}x})dx\]

\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{(\sin x)+\int\limits_{0}^{\dfrac{\pi }{2}}{(-\sin x)}{{\cos }^{2}}x}dx\]

Let \[I=A+B\] where,

\[A=\int\limits_{0}^{\dfrac{\pi }{2}}{(\sin x)}\]

\[B=\int\limits_{0}^{\dfrac{\pi }{2}}{(-\sin x)}{{\cos }^{2}}xdx\]

In B, let \[\cos x\]=t. So, \[(-\sin x)dx=dt\]

\[B=\int{{{t}^{2}}dt}=\dfrac{{{t}^{3}}}{3}=\dfrac{{{\cos }^{3}}x}{3}\]

Now,

\[I=-\cos x+\dfrac{{{\cos }^{3}}x}{3}\]

Now, we will apply the given limits,

\[\left[ -\cos x+\dfrac{{{\cos }^{3}}x}{3} \right]_{0}^{\dfrac{\pi }{2}}\]

\[\begin{align}

  & =\left[ -\cos \dfrac{\pi }{2}+\dfrac{{{\cos }^{3}}\dfrac{\pi }{2}}{3} \right]-\left[ -\cos 0+\dfrac{{{\cos }^{3}}0}{3} \right] \\

 & =0-\left[ -1+\dfrac{1}{3} \right]=1-\dfrac{1}{3}=\dfrac{2}{3} \\

\end{align}\]

So, the answer is \[\dfrac{2}{3}\].

Note: While solving the definite integral students must carefully apply the limits after simplifying the given integral and do calculations carefully to avoid error in answer.