Evaluate the integrals. ∫xsin⁡x2cos8x2dx

Question

Evaluate the integrals.  ∫xsin⁡x2cos8x2dx

Bertha Jordan · Accepted Answer

Step 1 Given- ∫xsin⁡x2cos8x2dx 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=∫xsin⁡x2cos8x2dx Now substituting cos⁡x2=t and differentiating both sides w.r.t. x, we get, cos⁡x2=t −sin⁡x2⋅2xdx=dt xsin⁡x2dx=−dt2 So, from the above integral, we can write as, I=∫t8⋅(−dt2) =−12⋅∫t8dt =−12⋅t99+C =−t918+C Step 3 As previously, we have substituted cos⁡x2=t, so the solution of the integeral can be written as, I=−(cos⁡x2)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 ∫xsin⁡x2cos8x2dx is −cos9x218+C, where C is an arbitrary constant.

Neil Dismukes · Accepted Answer

∫xcos8(x2)sin⁡(x2)dx  =−12∫u8du  ∫u8du  =u99  −12∫u8du  =−u918  =−cos9(x2)18  ∫xcos8(x2)sin⁡(x2)dx  =−cos9(x2)18+C

karton · Accepted Answer

∫x×sin⁡(x2)cos⁡(x2)8dx∫−t82dt−12×∫t8dt−12×t99−12×cos⁡(x2)99−cos⁡(x2)918

Evaluate the integrals. \int x\sin x^{2}\cos^{8}x^{2}dx

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