Q5  Find the integrals of the functions \sin ^3 x \cos ^ 3 x

Answers (1)
M manish

I = \int \sin^3x.\cos^3x\ dx

rewrite the integral as follows

    \\=\int cos^3x.sin^2x.\sin x\ dx\\ =\int cos^3x(1-\cos^2x)\sin x\ dx
 Let   \cos x = t \Rightarrow \sin x dx =-dt

   \\=-\int t^3(1-t^2)dt\\ =-\int(t^3-t^5)dt\\ =-[\frac{t^4}{4}]+[\frac{t^6}{6}] +C\\ =\frac{\cos^6x}{6}-\frac{cos^4x}{4}+C......(replace the value of t as cos\ x)

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