Evaluate the integral $$I=\int_0^{\pi/2} \sin x(f'(\cos x)-2)\, dx$$ when $f(0)=1$ and $f(1)=5$.

I'm stuck in this question especially $f'(\cos x)$ don't know it should be $\sin x$ and the integral will be $\sin x(\sin x-2)$ or what

need help please


Hint: Let $t = \cos x \Rightarrow dt = - \sin x \, dx$. Then

$$I = \int_1^0 -(f'(t)-2) \, dt = \int_0^1 f'(t)-2 \, dt.$$

Solution: We have

$$I = \int_0^1 f'(t)-2 \, dt = f(1) - f(0) -2t]_0^1 = 5 - 1 -2 = 2.$$

by the fundamental theorem of calculus.

  • $\begingroup$ f'(t) = ???? need that to solve $\endgroup$ – user155971 Sep 8 '15 at 5:06
  • 1
    $\begingroup$ What does the fundamental theorem of calculus say about the integral $\int_a^b f'(t) \, dt$? $\endgroup$ – MathMajor Sep 8 '15 at 5:08
  • $\begingroup$ f(b)-f(a) I think that $\endgroup$ – user155971 Sep 8 '15 at 5:09
  • $\begingroup$ @user155971 That is correct. Keep going. I have edited my answer to include the full solution, but don't look at it until you try it yourself. $\endgroup$ – MathMajor Sep 8 '15 at 5:09

Inspection reveils that $$\phi(x):=-f(\cos x)+2\cos x$$ is a primitive of the integrand. The value of the integral is therefore given by $$\phi\left({\pi\over2}\right)-\phi(0)=\bigl(-f(0)+0\bigr)-\bigl(-f(1)+2\bigr)=2\ .$$


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