First slide

Methods of integration

Question

$\int \sin^{3} (2 x + 1) dx$ is equal to

Moderate

A

$- \frac{1}{2} \cos (2 x + 1) + \frac{\cos^{3} (2 x + 1)}{6} + C$
B

$- \cos (2 x + 1) + \frac{1}{3} \cos^{3} (2 x + 1) + C$
C

$\frac{1}{2} \cos (2 x + 1) - \frac{1}{3} \cos^{3} (2 x + 1) + C$
D

None of the above

Solution

$\begin{array}{l} \int \sin^{3} (2 x + 1) dx = \int \sin^{2} (2 x + 1) \cdot \sin (2 x + 1) dx \\ = \int [1 - \cos^{2} (2 x + 1)] \sin (2 x + 1) dx [∵ \sin^{2} x = 1 - \cos^{2} x] \\ Let \cos (2 x + 1) = t \Rightarrow - 2 \sin (2 x + 1) dx = dt \\ \Rightarrow \sin (2 x + 1) dx = - \frac{dt}{2} \\ ∴ \int \sin^{3} (2 x + 1) dx = \frac{- 1}{2} \int (1 - t^{2}) dt = \frac{- 1}{2} (t - \frac{t^{3}}{3}) + C \\ = \frac{- 1}{2} [\cos (2 x + 1) - \frac{\cos^{3} (2 x + 1)}{3}] + C \\ = \frac{- \cos (2 x + 1)}{2} + \frac{\cos^{3} (2 x + 1)}{6} + C \end{array}$

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