First slide

Methods of integration

Question

If $I_{n} = \int \tan^{n} xdx$ then $(n - 1) (I_{n} + I_{n - 2})$ is equal to

Moderate

A

$\tan^{n} x$
B

$\cot^{n} x$
C

$\tan^{n - 1} x$
D

$\cot^{n - 1} x$

Solution

Here,
$\begin{array}{l} I_{n} = \int \tan^{n} xdx = \int \tan^{n - 2} {xtan}^{2} xdx \\ = \int \tan^{n - 2} (\sec^{2} x - 1) dx = \int \tan^{n - 2} x \cdot \sec^{2} xdx - \int \tan^{n - 2} xdx \end{array}$
$\begin{matrix} = \int \tan^{n - 2} {xsec}^{2} xdx - I_{n - 2} \\ I_{n} + I_{n - 2} = \frac{\tan^{n - 1} x}{n - 1} \end{matrix}$
Hence, $(n - 1) (I_{n} + I_{n - 2}) = \tan^{n - 1} x$

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