Choose the correct option regarding the following definite integrals (i) $\int_0^{\pi / 2} \sin ^m(x) \cos…
Choose the correct option regarding the following definite integrals
(i) $\int_0^{\pi / 2} \sin ^m(x) \cos (x) d x=\frac{1}{m+1}$
(ii) $\int_0^{\pi / 2} \sin (x) \cos ^n(x) d x=\frac{1}{n+1}$
(i) is true, (ii) is false
(i) is false, (ii) is true
Both (i) and (ii) are false
Both (i) and (ii) are true
Solution
(i) $\int_0^{\pi / 2} \sin ^m x \cdot \cos x d x$
Put, $\sin x=t$
$
\begin{aligned}
\cos x d x & =d t=\int_0^{\pi / 2} t^m d t \\
& =\left[\frac{t^{m+1}}{m+1}\right]_0^{\pi / 2}=\left[\frac{(\sin x)^{m+1}}{m+1}\right]_0^{\pi / 2}
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{m+1}\left[(\sin x)^{m+1}\right]_0^{\pi / 2} \\
& =\frac{1}{m+1}\left[\sin ^{m+1}\left(\frac{\pi}{2}\right)-\sin ^{m+1} \cdot(0)\right] \\
& =\frac{1}{m+1}[(1)-(0)]=\frac{1}{m+1}
\end{aligned}
$
(ii) $\int_0^{\pi / 2} \sin x \cdot \cos ^n(x)=d x$
$
\begin{aligned}
\cos x & =t \\
-\sin x d x & =d t \\
\sin x \cdot d x & =-d t \\
\int_0^{\pi / 2} t^n \cdot(-d t) & =-\int_0^{\pi / 2} t^n \cdot d t \\
= & -\left[\frac{t^{n+1}}{n+1}\right]_0^{\pi / 2}=-\left[\frac{\cos x^{n+1}}{n+1}\right]_0^{\pi / 2} \\
& =\frac{-1}{n+1}\left[\cos ^{n+1} x\right]_0^{\pi / 2}
\end{aligned}
$
$
\begin{aligned}
& =\frac{-1}{n+1}\left[\cos ^{n+1} \frac{\pi}{2}-\cos ^{n+1}(0)\right] \\
& =\frac{-1}{n+1}[0-1]=\frac{1}{n+1}
\end{aligned}
$
Hence, option (4) is correct