Question

How to start integrating those integrals:
1)Integral of dx/(sqrt of 4 (sin^3xcos^5x)
2)Integral of dx/(sqrt(5+2x+x^2))^3
3)e^arctgx/(1+x^2)(sqrt (1+x^2))dx
4)xe^arctgx/(1+x^2)^3/2 dx
5)2x^7 -x^5+2x^3-x+1/cos^2x dx (f or this one -limit:from -pi/3 to pi/3)

Answer 1

For clarity:
\[\int\frac{\mathrm{d}x}{\sqrt{4\sin^3x\cos^5x}}\]
\[\int\frac{\mathrm{d}x}{\left(\sqrt{5+2x+x^2}\right)^3}\]
\[\int \frac{e^{\arctan x}}{(1+x^2)\sqrt{1+x^2}}\,\mathrm{d}x\]
\[\int \frac{x e^{\arctan x}}{(1+x^2)^{3/2}}\,\mathrm{d}x\]
\[\int_{-\pi/3}^{\pi/3}\frac{2x^7 -x^5+2x^3-x+1}{\cos^2x}\,\mathrm{d}x\quad\text{or}\\\quad
\quad\quad \int_{-\pi/3}^{\pi/3}\left(2x^7 -x^5+2x^3-x+\frac{1}{\cos^2x}\right)\,\mathrm{d}x~?\]
Are these correct?

Answer 2

Well... pending your approval, I'll assume everything above is correct. And for the last integral, I'll assume the latter is the correct - and also significantly easier - one. Might as well start off with that.

Since you're integrating over a symmetric interval, i.e. \(\displaystyle\int_{-a}^a\), you can make use of the fact that the first four terms of the integrand are odd. That is, for an odd function \(f(x)\) continuous over an interval \([-a,a]\), you have \(\displaystyle \int_{-a}^af(x)\,\mathrm{d}x=0\). This means the last integral is
\[\int_{-\pi/3}^{\pi/3}\left(2x^7 -x^5+2x^3-x+\frac{1}{\cos^2x}\right)\,\mathrm{d}x=\int_{-\pi/3}^{\pi/3}\sec^2x\,\mathrm{d}x\]

Answer 3

Might as well continue with the bottom-up trend. For the next integral,
\[\int \frac{x e^{\arctan x}}{(1+x^2)^{3/2}}\,\mathrm{d}x\]the \(\arctan x\) and factor of \(1+x^2\) in the denominator is a strong hint. If, say \(x=\tan u\) so that over a restricted domain \(u=\arctan x\), you have \(\mathrm{d}u=\dfrac{\mathrm{d}x}{1+x^2}\). The integral can then be rewritten as
\[\int\frac{e^u\tan u}{(1+\tan^2u)^{1/2}}\,\mathrm{d}u\]which is a bit more tractable.

Answer 4

The next one on the list uses the same logic as the previous one.

Answer 5

For the next one, completing the square is your friend:
\[x^2+2x+5=x^2+2x+4+1=(x+2)^2+1\]followed by a trig substitution of \(x+2=\tan u\), so that \(\mathrm{d}x=\sec^2u\,\mathrm{d}u\). The integral is then
\[\int\frac{\mathrm{d}x}{\left(\sqrt{5+2x+x^2}\right)^3}=\int\frac{\sec^2u}{\left(\sqrt{\tan^2u+1}\right)^3}\,\mathrm{d}u=\int\cos u\,\mathrm{d}u\][Note: the more pedantic user would suggest the final integral isn't identical to the previous one, and s/he'd be right. Behind the scenes, I'm really making the substitution \(\arctan(x+2)=u\). It's a tacit assumption we're making here - one that allows us to do what we do because we're restricting the domain to ensure that simplification and back-substituting yields a result that's easier on the eyes.]

Answer 6

Finally,
\[\begin{align*}\int\frac{\mathrm{d}x}{\sqrt{4\sin^3x\cos^5x}}&=\frac{1}{2}\int\frac{\mathrm{d}x}{\sqrt{\sin^3x\cos^5x}}\times\frac{\sec^4x}{\sec^4x}\\[1ex]
&=\frac{1}{2}\int\frac{\sec^4x}{\sqrt{\sin^3x\sec^3x}}\,\mathrm{d}x\\[1ex]
&=\frac{1}{2}\int\frac{\sec^2x\sec^2x}{\sqrt{\tan^3x}}\,\mathrm{d}x
\end{align*}\]I'll let you think about what the next step might be. The hint lies in the form of the numerator.

Answer 7

My mistake it is
>(2x^7 -x^5+2x^3-x+1)/cos^2x dx (for this one -limit:from -pi/3 to pi/3)

Answer 8

I'll think about next steps..👍

Answer 9

Actually, I take back what I said about one form being significantly easier than the other - I was thinking of the general antiderivative.

The same argument applies. \(\dfrac{x^{\text{odd positive integer}}}{\cos^2x}\) is an odd function, so the definite integral over \((-a,a)\) is still \(0\). The integral remains the same in this case.