∫sin( ax+b ) #160; cos( ax+b ) #160; #160; dx=∫ 1 2 #160; .2sin( ax+b ) #160; cos( ax+b ) #160; dx = 1 2 ∫sin 2( ax+b ) #160; ...

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Antiderivative

∫sinax+bcos a...

Question

$\int sin (a x + b) cos (a x + b) d x$ .

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Solution

$\int sin (a x + b) cos (a x + b) d x = \int \frac{1}{2} .2 sin (a x + b) cos (a x + b) d x$
$= \frac{1}{2} \int sin 2 (a x + b) [∵ sin 2 x = 2 sin x cos x]$
$= \frac{1}{2} \int sin (2 a x + 2 b) d x = - \frac{1}{2} \frac{cos (2 a x + 2 b)}{2 a} + c = \frac{- cos 2 (a x + b)}{4 a} + c$

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