integration of sin(ax+b) cos(ax+b) dx is my method and answer correct (ncert exercise 7 2 - Maths - Integrals

Question

integration of sin(ax+b) cos(ax+b) dx ?? is my method and answer
correct?? (ncert exercise 7 2 Q5)
ans - let cos(ax+b) = t
-asin(ax+b) dx = dt
=> 1/a X integration of t dt
= 1/2a X t2 + C
=1/2a X cos2 (ax+b) +C

Lalit Mehra · Accepted Answer

$\int \sin (a x + b) \cos (a x + b) ⅆ x Let \cos (a x + b) = t$

∴ – a sin (ax + b) dx = dt

$\Rightarrow \sin (a x + b) d x = - \frac{d t}{a}$

∴ ∫ sin (ax + b) cos (ax + b) dx

$= - \frac{1}{a} \int t ⅆ t = - \frac{1}{a} \times \frac{t^{2}}{2} + C = - \frac{1}{2 a} [\cos (a x + b)]^{2} + C = - \frac{1}{2 a} \cos^{2} (a x + b) + C$

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integration of sin(ax+b) cos(ax+b) dx ?? is my method and answer correct?? (ncert exercise 7.2 Q5) ans - let cos(ax+b) = t -asin(ax+b) dx = dt => 1/a X integration of t dt = 1/2a X t2 + C =1/2a X cos2 (ax+b) +C

integration of sin(ax+b) cos(ax+b) dx ?? is my method and answer correct?? (ncert exercise 7.2 Q5)

ans - let cos(ax+b) = t

-asin(ax+b) dx = dt

=> 1/a X integration of t dt

= 1/2a X t² + C

=1/2a X cos² (ax+b) +C