Question

How do you do this problem?

Answer 1

Answer:

$\int\limits^{\sqrt{\pi}}_0 {4x^3\cos(x^2)} \, dx=-4$

Step-by-step explanation:

So we have the integral:

$\int\limits^{\sqrt{\pi}}_0 {4x^3\cos(x^2)} \, dx$

As told, let's use u-substitution first and then use integration by parts.

For the u-substitution, we can let u to be equal to x². So:

$u=x^2$

Differentiate:

$du=2x\, dx$

We can rewrite our integral as:

$\int\limits^{\sqrt{\pi}}_0 {2x(2x^2)\cos(x^2)} \, dx$

Therefore, by making our u-substitution, our integral is now:

$\int\limits {2u\cos(u)} \, du$

We also need to change our bounds. Substitute them into u. So:

$u=\sqrt{\pi}^2=\pi\\u=(0)^2=0$

Therefore, our integral with our new bounds is:

$\int\limits^{\pi}_{0} {2u\cos(u)} \, du$

Now, let's use integration by parts. Integration by parts is given by:

$\int\limits {v}\, dy=vy-\int y\, dv$

(I changed the standard u to y because we are already using u).

Let's let v be 2u and let's let dy be cos(u). Thus:

$v=2u\\dv=2\,du$

And:

$dy=\cos(u)\\y=\sin(u)$

So, do integration by parts:

$=2u\sin(u)-\int \sin(u)2\,du$

Simplify:

$=2u\sin(u)-2\int \sin(u)\,du$

Evaluate the integral:

$=2u\sin(u)+2\cos(u)$

Now, use the bounds. So:

$(2(\pi)\sin(\pi)+2\cos(\pi))-(2(0)\sin(0)+2\cos(0))$

Evaluate:

$=(2\pi(0)+2(-1))-(0(0)+2(1))$

Simplify:

$=(-2)-(2)$

Subtract:

$=-4$

And we're done!

Answer 2

Answer:

Your answer is absolutely correct

Step-by-step explanation:

The work would be as follows:

$\int _0^{\sqrt{\pi }}4x^3\cos \left(x^2\right)dx,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=> 4\cdot \int _0^{\sqrt{\pi }}x^3\cos \left(x^2\right)dx\\\\\mathrm{Apply\:u-substitution:}\:u=x^2\\=> 4\cdot \int _0^{\pi }\frac{u\cos \left(u\right)}{2}du\\\\\mathrm{Apply\:Integration\:By\:Parts:}\:u=u,\:v'=\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\int \sin \left(u\right)du\right]^{\pi }_0$

$\int \sin \left(u\right)du=-\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0\\\\\mathrm{Simplify\:}4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0:\quad 2\left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0\\\\\mathrm{Compute\:the\:boundaries}:\quad \left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0=-2\\=> 2(-2) = - 4$

Hence proved that your solution is accurate.