What are the ways to solve this integral?

$$\int _{0}^{\frac{\pi}{2}}\frac{1}{5\sin x+3} dx.$$

I've tried the way of let $u=\tan \left( \frac{x}{2}\right)$ , but it's very complicated.

• – Lucian Mar 10 '15 at 9:50
• The Weierstrass substitution $u=tan(x/2)$ always works, but (as noted) is very complicated. So other methods (if they can be found) are useful to know. – GEdgar Mar 10 '15 at 13:34

As $\sin2u=\dfrac{2\tan u}{1+\tan^2u}$

$$\int_0^{\dfrac\pi2}\frac{dx}{5\sin x+3}=2\int_0^1\frac{dt}{10t+3(1+t^2)}$$

$$=2\int_0^1\frac{dt}{(3t+1)(t+3)}$$

$$=\frac23\int_0^1\frac{dt}{(t+1/3)(t+3)}$$

$$=\frac23\cdot\frac38\int_0^1\frac{(t+3)-(t+1/3)}{(t+1/3)(t+3)}dt$$

Can you tak it from here?

Alternatively, use $\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$

Then employ $\tan\dfrac x2=u\implies\cos u=\dfrac{1-u^2}{1+u^2}$

• I don't understand why 1/2. – Mathxx Mar 10 '15 at 9:17
• It's not equal to 2 – Mathxx Mar 10 '15 at 9:20
• @Mathxx, $\tan\dfrac x2=u\implies\sec^2\dfrac x2\dfrac{dx}2=du\implies dx=?$ as $\sec^2\dfrac x2=1+\tan^2\dfrac x2=1+u^2$ – lab bhattacharjee Mar 10 '15 at 9:22

If you want to avoid trigonometry, it's possible: you can substitute $u=5\sin x + 3$ so that $\mathrm{d}x = \frac{1}{5 \cos x} \mathrm{d}u$. Since you have $0 \leq x \leq \pi/2$ then $\cos x \geq 0$ and we obtain:

$$5\cos x = 5\sqrt{1 - \sin^2 x} = 5\sqrt{1 - \left(\frac{u-3}{5}\right)^2}=\sqrt{16 + 6u - u^2}$$

You may prefer to factorise this to $\sqrt{-(u+2)(u-8)}$. After changing the limits the integral has become:

$$\int _{0}^{\pi/2}\frac{1}{5\sin x+3} \mathrm{d}x = \int _{3}^{8}\frac{1}{u \sqrt{-(u+2)(u-8)}} \mathrm{d}u$$

How has this helped avoid integrating trig? After all, seeing a quadratic inside a square root is exactly the time that a lot of people will just use a trig substitution! But there is an alternative, which is to make an Euler substitution. In fact there are several which will work here, although Euler's first substitution will not because the leading coefficient of the quadratic under the square root is negative. The second substitution would work because the constant term in that quadratic is greater than zero, and the third substitution will work because the quadratic factorises (as above).

The analysis from this point is extremely similar to this answer. If we implemented Euler's third substitution as $\sqrt{-(u+2)(u-8)} = (u+2)t$ then we obtain, after a bit of algrebra, an integral than be solved easily by partial fractions:

$$\int_0^1 \frac{1}{4-t^2} \mathrm{d}t = \frac{1}{4} \int_0^1 \frac{1}{2+t} - \frac{-1}{2-t} \mathrm{d}t = \frac{1}{4} \bigg[ \log |2+t| - \log |2-t| \bigg]_0^1 =\frac{\log 3}{4}$$

Of course, even the final integral might have been easier with a (hyperbolic) trig substitution. But the question seemed to be seeking alternatives, and this is certainly one of them.

• This deserves a vote. – Mathxx Mar 10 '15 at 15:42

No it's not very complicated, proceed carefully: $$\scriptsize I=\int\frac{{\rm d}x}{3+5\sin x}=\int\frac{{\rm d}x}{3+5\left(\frac{2\tan(x/2)}{1+\tan^2(x/2)}\right)}=\int\frac{(1+\tan^2(x/2)){\rm d}x}{3(1+\tan^2(x/2))+10\tan (x/2)}=\int\frac{\sec^2(x/2){\rm d}x}{3\tan^2(x/2)+10\sin x+3}$$ Now, take $y=\tan(x/2)\implies {\rm d}y=\sec^2(x/2).(1/2).{\rm d}x$ using Chain Rule. $$I=\int\frac{2{\rm d}y}{3y^2+10y+3}=\int\frac{2{\rm d}y}{3(y+5/3)^2-16/3}$$ Now take $z=\sqrt3(y+5/3),{\rm d}z=\sqrt3{\rm d}y$: $$I=\frac2{\sqrt3}\int\frac{{\rm d}z}{z^2-(4/\sqrt3)^2}=\frac2{\sqrt3}.\frac1{2(4/\sqrt3)}\ln\left|\frac{z-4/\sqrt3}{z+4/\sqrt3}\right|+{\rm C}$$ So: $$\int\frac{{\rm d}x}{3+5\sin x}=\frac14\ln\left|\frac{3(\tan (x/2)+5/3)-4}{3(\tan (x/2)+5/3)+4}\right|+{\rm C}=\frac14\ln\left|\frac{3\tan (x/2)+1}{3\tan (x/2)+9}\right|+{\rm C}$$ Now just put the limits: $$\frac14\ln\left|\frac{3\tan (x/2)+1}{3\tan (x/2)+9}\right|\Bigg|_0^{\pi/2}=\frac14\ln\left|\frac{3+1}{3+9}\right|-\frac14\ln\left|\frac{0+1}{0+9}\right|=\frac14\ln\frac{4/12}{1/9}=\frac14\ln3\approx0.27465$$ Moral:

$$\text{You are just lazy.}$$