( ax + b ) ( ax + b )

.: Let I = ) ( ax + b )dx Put ( ax + b ) = t a ( ax + b )dx = dt I = a t dt = a 2 + C = 2a t 2 + C = 2a 2 ( ax + b ) + C Aliter : Put ( ax + b ) = t - a ( ax + b )dx = dt I = a t dt = a 2 + C = 2 ( ax + b ) 2a + C

class 12 maths integrals

$\sin \left( {ax + b} \right)\cos \left( {ax + b} \right)$

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#Integrals #NCERT #SA #Class 12 Maths

📘 Integrals NCERT,ex.7.2,Q.5,Page 304 SA

$\sin \left( {ax + b} \right)\cos \left( {ax + b} \right)$

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

.: Let$I = \int {\sin \left( {ax + b} \right)\cos \left( {ax + b} \right)dx}$

Put $\sin \left( {ax + b} \right) = t$ $\Rightarrow$ $a\cos \left( {ax + b} \right)dx = dt$

$\therefore$ $I = \cfrac{1}{a}\int t dt = \cfrac{1}{a} \cdot \cfrac{{{t^2}}}{2} + C = \cfrac{1}{{2a}}{t^2} + C = \cfrac{1}{{2a}}{\sin ^2}\left( {ax + b} \right) + C$

Aliter : Put $\cos \left( {ax + b} \right) = t$ $\Rightarrow$ $- a\sin \left( {ax + b} \right)dx = dt$

$\therefore$ $I = \cfrac{{ - 1}}{a}\int t dt = \cfrac{{ - 1}}{a} \cdot \cfrac{{{t^2}}}{2} + C = \cfrac{{ - {{\cos }^2}\left( {ax + b} \right)}}{{2a}} + C$

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