Evaluate the integrals. ∫xsin⁡x2cos8x2dx

petrusrexcs

petrusrexcs

Answered

2021-12-29

Evaluate the integrals.
xsinx2cos8x2dx

Answer & Explanation

Bertha Jordan

Bertha Jordan

Expert

2021-12-30Added 37 answers

Step 1
Given- xsinx2cos8x2dx
To find- The value of the above integral.
Concept Used- The above integral can be solved by substitution method.
Step 2
Explanation- Rewrite the given expression as,
I=xsinx2cos8x2dx
Now substituting cosx2=t and differentiating both sides w.r.t. x, we get,
cosx2=t
sinx22xdx=dt
xsinx2dx=dt2
So, from the above integral, we can write as,
I=t8(dt2)
=12t8dt
=12t99+C
=t918+C
Step 3
As previously, we have substituted cosx2=t, so the solution of the integeral can be written as,
I=(cosx2)918+C
=cos9x218+C, where C is arbitrary constant.
So, the solution of the integral is cos9x218+C.
Answer- Hence, the solution of the integral xsinx2cos8x2dx is cos9x218+C, where C is an arbitrary constant.

Neil Dismukes

Neil Dismukes

Expert

2021-12-31Added 37 answers

xcos8(x2)sin(x2)dx
=12u8du
u8du
=u99
12u8du
=u918
=cos9(x2)18
xcos8(x2)sin(x2)dx
=cos9(x2)18+C
karton

karton

Expert

2022-01-04Added 439 answers

x×sin(x2)cos(x2)8dxt82dt12×t8dt12×t9912×cos(x2)99cos(x2)918

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