$\int \sin ^5 x \cdot \cos ^5 x d x=$
$\int \sin ^5 x \cdot \cos ^5 x d x=$
- $\frac{\cos ^6 x}{60}\left(6 \sin ^4 x+3 \sin ^2 x+1\right)+c$
- $-\frac{\sin ^6 x}{60}\left(6 \cos ^4 x+3 \cos ^2 x+1\right)+c$
- $-\frac{\cos ^6 x}{60}\left(6 \sin ^4 x+3 \sin ^2 x+1\right)+c$
- $\frac{\sin ^6 x}{60}\left(6 \cos ^4 x+3 \cos ^2 x+1\right)+c$
Solution
$
\begin{aligned}
I & =\int \sin ^5 x \cos ^5 x d x \\
& =\int \cos ^5 x \sin ^4 x \sin x d x \\
& =\int \cos ^5 x\left(1-\cos ^2 x\right)^2 \sin x d x
\end{aligned}
$
Let $\cos x=t \Rightarrow-\sin x d x=d t$
$
\begin{aligned}
& \text { So, } I=\int t^5\left(1-t^2\right)^2(-d t) \\
& =-\int t^5\left(t^4+1-2 t^2\right) d t=-\int\left(t^9-2 t^7+t^5\right) d t \\
& =-\left[\frac{t^{10}}{10}-2 \frac{t^8}{8}+\frac{t^6}{6}\right]+c=-\frac{t^6}{60}\left[6 t^4-15 t^2+10\right]+C \\
& =-\frac{\cos ^6 x}{60}\left[6 \cos ^4 x-15 \cos ^2 x+10\right]+C \\
& =-\frac{\cos ^6 x}{60}\left[6\left(1-\sin ^2 x\right)^2-15\left(1-\sin ^2 x\right)+10\right]+c
\end{aligned}
$
[put the value of $t$ ]
$
\begin{aligned}
&=-\frac{\cos ^6 x}{60}\left[6 \sin ^4 x-12 \sin ^2 x+\right.-15 \\
&\left.+15 \sin ^2 x+10\right]+C \\
&=-\frac{\cos ^6 x}{60}\left[6 \sin ^4 x+3 \sin ^2 x+1\right]+C
\end{aligned}
$
Hence, option (c) is correct
Asked in: AP EAMCET 2019 (20 Apr Shift 2)
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