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Problem:

Integrate : $\displaystyle \int^\pi_0\hspace{1mm}\sin^2t\cos^4t~dt$

What I tried :

  1. Use the pythagorean identity $\cos^2t=1-\sin^2t$ to rewrite the problem as $$\int^\pi_0\hspace{1mm}\sin^2t\hspace{1mm}(1-\sin^2t)\cos^2t\,dt$$

  2. Distribute $\sin^2t$ to rewrite the problem as

    $$\int^\pi_0\hspace{1mm}(\sin^2t-\sin^4t)\cos^2t \, dt$$

  3. let $u = \sin^2t$ ; $du=2\cos^2t\hspace{1mm}dt$

  4. Make the u substitution to rewrite the problem as

    $$\frac 1 2 \int^{\sin^2\pi}_{\sin^20}\hspace{1mm}u - u^2\,du$$

When evaluated the whole thing obviously just goes to zero. When I look in the back of the book the correct answer is $\hspace{2mm}\pi/16$

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    $\begingroup$ You substituted $u = \sin^2 t$. So $du = 2 \sin t \cos t \ dt$. $\endgroup$
    – jonsno
    Commented Sep 21, 2017 at 3:19

1 Answer 1

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Hint:) Use identities $$\sin^2t=\dfrac{1-\cos2t}{2}~~~~,~~~~\cos^2t=\dfrac{1+\cos2t}{2}$$ and reduce powers.

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  • $\begingroup$ I have qualms about the use of the word "hint" for this sort of thing. $\endgroup$ Commented Sep 21, 2017 at 3:30
  • $\begingroup$ @MichaelHardy I can delete this qualms thing if you want! $\endgroup$
    – Nosrati
    Commented Sep 21, 2017 at 3:33
  • $\begingroup$ @MyGlasses Thanks, That does get me toward the right answer, but where did I go wrong in the substitution I made? $\endgroup$
    – richbai90
    Commented Sep 21, 2017 at 3:38
  • $\begingroup$ @MyGlasses Nevermind I borked the derivative. THANKS $\endgroup$
    – richbai90
    Commented Sep 21, 2017 at 3:41
  • $\begingroup$ You made a mistake in $du$ as @samjoe said in his comments, anyway this substitution isn't applicable since it need a term of $cos$ with power $1$ only. $\endgroup$
    – Nosrati
    Commented Sep 21, 2017 at 3:41

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