Evaluate the integrals  \int sin^3x\;cos^6x\;dx

Evaluate the integrals \( \int sin^3x\;cos^6x\;dx\)

Asked by Sakshi | 1 year ago |  159
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1 Answer

Solution :-

\( \int sin^3x\;cos^6x\;dx\)

=\( ∫sin^2x\;sinx\;cos^6xdx\)

=\( ∫(1−cos^2x)sinx\;cos^6x\;dx\)

=\( ∫(cos^6x−cos^8x)sinx\;dx\)

Let t = cosx

= dt =−sinxdx

\(- \int (t ^6−t ^8)dt\)

\( \dfrac{cos^9x}{9}- \dfrac{cos^7x}{7}+c\) where t=cosx

Answered by Sakshi | 1 year ago
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