# Compute \int_0^{\sqrt {\frac {\pi}{2}}} \int_{x^2}^{\frac{\pi}{2}} 6x \sin y \ dy \ dx

## Question:

Compute {eq}\int_0^{\sqrt {\frac {\pi}{2}}} \int_{x^2}^{\frac{\pi}{2}} 6x \sin y \ dy \ dx {/eq}

## Double Integral In Calculus:

Iterated integrals are used when there are two or more variables present in the expression of integration. We are given a double integral, which is used for a function of two variable over a two-dimensional region on {eq}xy {/eq} space.

To solve this problem, we'll use common integrals. Next, use u-substitution, which is one of the most common rules to compute intergral.

## Answer and Explanation:

We are given:

{eq}\displaystyle \int_0^{\sqrt {\frac {\pi}{2}}} \int_{x^2}^{\frac{\pi}{2}} 6x \sin y \ dy \ dx {/eq}

Now, solving the innermost part of the integral, we will integrate in terms of {eq}x {/eq}:

{eq}= \displaystyle \int_{x^2}^{\frac{\pi}{2}} 6x \sin y \ dy {/eq}

Take the constant out:

{eq}=\displaystyle 6x \int_{x^2}^{\frac{\pi}{2}} \sin y \ dy {/eq}

{eq}=\displaystyle \left[ -6x\cos y \right]_{x^2}^{\frac{\pi}{2}}\\ {/eq}

Compute the boundaries:

{eq}=\displaystyle -6x \cos \frac{\pi}{2}+6x \cos x^2 {/eq}

{eq}=\displaystyle 6x \cos x^2 {/eq}

Now, the integral will be:

{eq}\displaystyle \int_0^{\sqrt {\frac {\pi}{2}}} 6x \cos x^2 \mathrm{d}x {/eq}

Apply u-substitution {eq}u = x^2 \rightarrow \ du = 2 x \ dx {/eq}, the limits will also change. The upper limit will be {eq}\sqrt {\frac {\pi}{2}} \rightarrow \frac {\pi}{2} {/eq} , the lower limit will be {eq}0 \rightarrow 0 {/eq}

{eq}=\displaystyle 3 \int_0^{\frac {\pi}{2}} \cos u \mathrm{d}u {/eq}

Take the constant out:

{eq}=\displaystyle 3\left[ \sin u \right]_0^{\frac {\pi}{2}} {/eq}

{eq}=\displaystyle 3\sin \frac{\pi}{2} - \sin 0 {/eq}

{eq}=\displaystyle 3 {/eq}

Therefore the solution is:

{eq}\displaystyle {\boxed{\int_0^{\sqrt {\frac {\pi}{2}}} \int_{x^2}^{\frac{\pi}{2}} 6x \sin y \ dy \ dx=3}} {/eq}

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from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14