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If $I_n=\int \sin ^n x d x$, then $n I_n-(n-1) I_{n-2}$ equals
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Verified Answer
The correct answer is:
$-\sin ^{n-1} x \cos x$
We know that, if
$\begin{aligned}
& I_n=\int \sin ^n x d x \text {, then } \\
& I_n=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} I_{n-2}
\end{aligned}$
where $n$ is a positive integer.
$\Rightarrow \quad n I_n-(n-1) I_{n-2}=-\sin ^{n-1} x \cos x$
$\begin{aligned}
& I_n=\int \sin ^n x d x \text {, then } \\
& I_n=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} I_{n-2}
\end{aligned}$
where $n$ is a positive integer.
$\Rightarrow \quad n I_n-(n-1) I_{n-2}=-\sin ^{n-1} x \cos x$
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