CBSE Class 12 Math 2013 Solved Paper

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Question : 17 of 29
 
Marks: +1, -0
Evaluate: ∫ sin(xa)sin(x+a) dx
OR
Evaluate: ∫ 5x21+2x+3x2 dx
Solution:
sin(xa)sin(x+a) dx
Let (x + a) = t ⇒ dx = dt
∴ I = ∫ sint2asint dt
= ∫ sintcos2acostsin2asint dt
= ∫ cos 2a - cot t sin 2a dt
= cos 2a t - sin 2a log |sin t| + C
= cos 2a x + a - sin 2a log |sin (x + a)| + C
OR
5x21+2x+3x2 dx
= 5 ∫ x251+2x+3x2 dx
= 566x1251+2x+3x2 dx
= 566x+212521+2x+3x2 dx
= 566x+22251+2x+3x2 dx
= 566x+21+2x+3x2 - 56×22513[(x+13)2+29] dx
= 56 log |1 + 2x + 3x2| - 1191(x+13)2+29 dx
= 56 log |1 + 2x + 3x2| - 119×32 tan1(x+13)23 + C
= 56 log |1 + 2x + 3x2| - 1132 × tan1(3x+12) + C
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