How do you do this problem?

Answer:

Your answer is absolutely correct

Step-by-step explanation:

The work would be as follows:

[tex]\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\\\\[/tex]

[tex]\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[/tex]

Hence proved that your solution is accurate.

Answer:

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

Step-by-step explanation:

So we have the integral:

[tex]\int\limits^{\sqrt{\pi}}_0 {4x^3\cos(x^2)} \, dx[/tex]

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:

[tex]u=x^2[/tex]

Differentiate:

[tex]du=2x\, dx[/tex]

We can rewrite our integral as:

[tex]\int\limits^{\sqrt{\pi}}_0 {2x(2x^2)\cos(x^2)} \, dx[/tex]

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

[tex]\int\limits {2u\cos(u)} \, du[/tex]

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

[tex]u=\sqrt{\pi}^2=\pi\\u=(0)^2=0[/tex]

Therefore, our integral with our new bounds is:

[tex]\int\limits^{\pi}_{0} {2u\cos(u)} \, du[/tex]

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

[tex]\int\limits {v}\, dy=vy-\int y\, dv[/tex]

(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:

[tex]v=2u\\dv=2\,du[/tex]

And:

[tex]dy=\cos(u)\\y=\sin(u)[/tex]

So, do integration by parts:

[tex]=2u\sin(u)-\int \sin(u)2\,du[/tex]

Simplify:

[tex]=2u\sin(u)-2\int \sin(u)\,du[/tex]

Evaluate the integral:

[tex]=2u\sin(u)+2\cos(u)[/tex]

Now, use the bounds. So:

[tex](2(\pi)\sin(\pi)+2\cos(\pi))-(2(0)\sin(0)+2\cos(0))[/tex]

Evaluate:

[tex]=(2\pi(0)+2(-1))-(0(0)+2(1))[/tex]

Simplify:

[tex]=(-2)-(2)[/tex]

Subtract:

[tex]=-4[/tex]

And we're done!

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