Q115 / 160 · TS EAMCET 2009 · Mathematics

If $I_n=\int \sin ^n x d x$, then $n I_n-(n-1) I_{n-2}$ equals

  • A
    $\sin ^{n-1} x \cos x$
  • B
    $\cos ^{n-1} x \sin x$
  • Correct answer
    C
    $-\sin ^{n-1} x \cos x$
  • D
    $-\cos ^{n-1} x \sin x$

Solution

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. $\Rightarrow \quad n I_n-(n-1) I_{n-2}=-\sin ^{n-1} x \cos x$

Hence, the correct answer is option: 'C'.

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