First slide

Methods of integration

Question

$Evaluate \int \sin^{3} x \cos^{5} x d x$

Easy

A

$- \frac{\cos^{5} x}{5} + \frac{\cos^{7} x}{7} + c$
B

$\frac{\cos^{6} x}{6} - \frac{\cos^{8} x}{8} + c$
C

$- \frac{\cos^{6} x}{6} + \frac{\cos^{8} x}{8} + c$
D

None of these

Solution

$H e r e, p o w e r s o f b o t h \cos x a n d \sin x a r e o d d p o s i t i v e i n t e g e r s;$
$t h e r e f o r e, p u t z = \cos x o r z = \sin x b u t t h e p o w e r o f \cos x i s g r e a t e r .$
$T h e r e f o r e, i t i s c o n v e n i e n t t o p u t z = \cos x$
$\begin{array}{l} Let z = \cos x . Then d z = - \sin x d x \\ \int \sin^{3} x \cos^{5} x d x \\ = \int \sin^{2} x \cos^{5} x \sin x d x \\ = \int (1 - \cos^{2} x) \cos^{5} x \sin x d x \\ = \int (1 - z^{2}) z^{5} (- d z) = - \int (z^{5} - z^{7}) d z \\ = - (\frac{z^{6}}{6} - \frac{z^{8}}{8}) + c = - \frac{\cos^{6} x}{6} + \frac{\cos^{8} x}{8} + c \end{array}$

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