Subjective Type

Integrate the function $$\sin (ax + b) \cos (ax + b)$$

Solution

$$\displaystyle \sin (ax+b)\cos (ax+b)=\frac {2\sin (ax+b)\cos (ax+b)}{2}=\frac {\sin 2(ax+b)}{2}$$
Put $$2(ax+b)=t$$
$$\Rightarrow 2adx=dt$$
$$\displaystyle =>\int \frac {\sin 2(ax+b)}{2}dx=\frac {1}{2}\int \frac {\sin t dt}{2a}$$
$$\displaystyle =\frac {1}{4a}[-\cos t]+C$$
$$\displaystyle =\frac {-1}{4a}\cos 2(ax+b)+C$$

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