First slide

Evaluation of definite integrals

Question

$\int \frac{\sin^{6} x}{\cos^{8} x} d x =$

Moderate

A

$\frac{\tan^{4} x}{4} + c$
B

$\frac{\tan^{16} x}{16} + c$
C

$\frac{\tan^{7} x}{7} + c$
D

$\frac{\tan^{8} x}{8} + c$

Solution

$\int \frac{\sin^{6} x}{\cos^{8} x} d x$
$\begin{array}{l} = \int \frac{\sin^{6} x}{\cos^{6} x} \times \frac{1}{\cos^{2} x} d x \\ = \int \tan^{6} x \sec^{2} x d x \\ = \frac{\tan^{7} x}{7} + c \end{array}$

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