$$\int_0^{\pi/2} \! \frac{\sin x\cos x}{(4-\sin^2 x)^2} \:\text{d}x$$

set $u = 4-\sin^2 x$, therefore $du = -2 \sin x \cos x \text{d}x $

$$-\frac{1}{2} \int u^{-1/2} \text{d}u $$

Change the range of integration from $0$ to $4$ and $\pi/2$ to $3$, and change upper and lower bounds to $4$ and $3$ respectively.

Now the problem reads:

$$-\frac{1}{2}\int_3^4 u^{-1/2}$$

$\int u^{-1/2} = 2(u^{1/2)}$

Multiply $2(u^{1/2})$ by $-1/2$ and Substitute in $4-\sin^2 x $ into $u$.

I got to this point and came up with the solution that the definite integral of the problem is solved to be $- \sqrt{\sin^2 x -4} + C$

After plugging in $4$ and $3$ the solution came to be $0.5528351602...$

I put it in to the wolfram alpha and it shows the answer to be $\sqrt{u}$, which is $\sqrt{sin^2(x)-4)} + C$

I need some explanation to why multiplying $-1$ to the integrand, in this case, is correct?

  • 2
    $\begingroup$ I fixed the LaTeX for you, but try to learn it yourself $\endgroup$ – Jack Oct 30 '14 at 3:56

You made a mistake in your u-substitution:

$$\text{Here's the interal}:\int_{0}^{\frac{\pi}{2}} \frac{\sin(x)\cos(x)}{(4-\sin^2(x))^2}dx $$

$$\text{Here's your u-sub}: u = 4-\sin^2{(x)}, \hspace{3mm}du = -2\sin(x)\cos(x)$$

Substituting this in should give you:

$$-\frac{1}{2} \int \frac{du}{u^2} = \frac{1}{2u}$$

Back-substituting this in gives you: $$\frac{1}{8-2\sin^2(x)}\bigg|_{0}^{\frac{\pi}{2}} = \frac{1}{6} - \frac{1}{8}.$$

Then again, did you intent to write the original problem as:

$$\int_{0}^{\frac{\pi}{2}} \frac{\sin(x)\cos(x)}{\sqrt{4-\sin^2(x)}}dx \text{?}$$


So the rule is

if f is integrable on [a,b]

then integral of f(x) dx from b to a is equal to - integral of f(x) dx from a to b.


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