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Question

Mathematics Question on integral

Find the integrals of the function: sin3x cos3xsin^3 x\, cos^3 x

Answer

The correct answer is: =cos6x6−cos4x4+C= \frac{cos^6x}{6}-\frac{cos^4x}{4}+C
Let I=∫sin3x cos3x.dxI = ∫sin^3 x\, cos^3 x.dx
=∫cos3x.sin2x.sinx.dx= ∫cos^3 x. sin^2 x. sin x. dx
=∫cos3x(1−cos2x)sinx.dx= ∫cos^3 x(1-cos^2 x)sin x.dx
Let cosx=1cos x = 1
⇒−sinx.dx=dt⇒ -sin x.dx = dt
⇒I=−∫t3(1−t2)dt⇒ I = - ∫t^3(1-t^2)dt
=−∫(t3−t5)dt= -∫(t^3-t^5)dt
=−[t44−t66]+C= -[{\frac{t^4}{4}-\frac{t^6}{6}}]+C
=−[cos4x4−cos6x6]+C= - [{\frac{cos^4x}{4}-\frac{cos^6x}{6}}]+C
=cos6x6−cos4x4+C= \frac{cos^6x}{6}-\frac{cos^4x}{4}+C