Integrate: ∫x2(xcos⁡x−sin⁡x)(xsin⁡x+cos⁡x)dx

Question

Integrate:  ∫x2(xcos⁡x−sin⁡x)(xsin⁡x+cos⁡x)dx

Kayla Kline · Accepted Answer

My solution:  Let I=∫x2(xsin⁡x+cos⁡x)⋅(xcos⁡x−sin⁡x)dx  So  I=∫tan2θ⋅sec2θ(tan⁡θ⋅sin⁡(tan⁡θ)+cos⁡(tan⁡θ))×(tan⁡θ⋅cos⁡(tan⁡θ)−sin⁡(tan⁡θ))dθ  So  I=∫tan2θ⋅sec2θcos2θ(sin⁡θ⋅sin⁡(tan⁡θ)+cos⁡(tan⁡θ)⋅cos⁡θ)  So  I=∫2tan2θ2cos⁡(θ−tan⁡θ)⋅sin⁡(θ−tan⁡θ)dθ=∫2tan2θsin⁡(2θ−2tan⁡θ)dθ  Now Let (2θ−2tan⁡θ)=u, then (2−2sec2θ)dθ=du⇒2tan2θdθ=−du  So  I=−∫1sin⁡udu=−∫csc⁡udu  =−ln⁡tan⁡(u2)+C  =−ln⁡tan⁡(θ−tan⁡θ)+C  So  I=ln|cos⁡(θ−tan⁡θ)sin⁡(θ−tan⁡θ)|+C  =ln|cos⁡θ⋅(tan⁡θ)+sin⁡θ⋅sin⁡(tan⁡θ)sin⁡θ⋅sin⁡(tan⁡θ)|+C  So

Heather Fulton · Accepted Answer

Another Solution:  Let I=∫x2(xsin⁡x+cos⁡x)⋅(xcos⁡x−sin⁡x)dx  So I=∫2x2(x2−1)sin⁡2x+2xcos⁡2xdx  Now (x2−1)sin⁡2x+2xcos⁡2x=(x2−1)2+(2x)2⋅[(x2−1x2−1)sin⁡2x+(2xx2+1)cos⁡2x]  Now we can write (x2−1)sin⁡2x+2xcos⁡2x=(x2+1)⋅sin⁡(2x+α)  Where (x2−1x2+1)=cos⁡α and (2xx2+1)=sin⁡α  and so tan⁡α=(2xx2−1)⇒α=tan−1(2xx2−1)  So Integral I=∫csc⁡(2x+α)⋅(2x2x2+1)dx  Now let (2x+α)=t⇒[2x+tan−1(2xx2−1)]=t, then (2x2x2+1)dx=dt  So integral I=∫csc⁡tdt=ln|tan⁡t2|+C=ln|tan⁡(x+α2)|+C

star233 · Accepted Answer

HINT : Rewrite the integrand x2(xcos⁡x−sin⁡x)(xsin⁡x+cos⁡x) as xcos⁡xxsin⁡x+cos⁡x+xsin⁡xxcos⁡x−sin⁡x then sin⁡x+xcos⁡x−sin⁡xxsin⁡x+cos⁡x+cos⁡x+xsin⁡x−cos⁡xxcos⁡x−sin⁡x Now let u=xsin⁡x+cos⁡x and v=xcos⁡x−sin⁡x

Integrate: \int\frac{x^2}{(x\cos x-\sin x)(x\sin x+\cos x)}dx

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Answer & Explanation

New Questions in Calculus 1