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a
12logesinx−(x+1)cosx−1sinx−(x+1)cosx+1+C
b
12tan−1{sinx−(x+1)cosx}+C
c
12sin−1{sinx−(x+1)cosx}+C
d
12sin−1(cosx+sinx)+C
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Correct option is A
I=∫(1+x)sin xdx(x+1)2cos2x−cos2 x−2(1+x)sin xcos xI=(1+x)sinxdx(x+1)2cos2x+sin2 x−2(1+x)sin xcos x−1I=(1+x)sinx(sinx−(x+1)cosx)2−1dx Put sinx−(x+1)cosx=t⇒sin x(x+1)dx=dt∴ I=∫dtt2−1=12log t−1t+1+C=12loge sin x−(x+1)cos x−1sin x−(x+1)cos x+1+CSimilar Questions
is equal to
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