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a
−cos5x5+cos7x7+c
b
cos6x6-cos8x8+c
c
−cos6x6+cos8x8+c
d
None of these
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Correct option is C
Here, powers of both cos x and sin x are odd positive integers; therefore, put z = cos x or z = sin x but the power of cos x is greater. Therefore, it is convenient to put z= cos x Let z=cosx . Then dz=−sinxdx∫sin3xcos5xdx=∫sin2xcos5xsinxdx=∫1−cos2xcos5xsinxdx=∫1−z2z5(−dz)=−∫z5−z7dz=−z66−z88+c=−cos6x6+cos8x8+cSimilar Questions
is equal to
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