Finding $\int\frac{\sin x+\tan x}{\cos x+\csc x}dx$

Question

Finding $$\displaystyle \int\frac{\sin x+\tan x}{\cos x+\csc x}\ dx$$

what I have tried

$$\Lambda =\int\frac{\sin^2 x(1+\cos x)}{\cos x(\sin x\cos x+1)}\ dx$$

$$\Lambda=\int\frac{\sin^4 x}{\cos x(1-\cos x)(\sin x\cos x+1)}\ dx$$

How do I solve it? Help please.

Answer 1

Hint: Use the so called Weierstrass Substitution: $$\sin(x)=\frac{2t}{1+t^2}$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$ $$dx=\frac{2dt}{1+t^2}$$

Answer 2

Hint: $$\text{Define }I=\underbrace{\int\dfrac{\tan x}{\cos x+\csc x}\mathrm dx}_{I_1}+\underbrace{\int\dfrac{\sin x}{\cos x+\csc x}\mathrm dx}_{I_2}$$

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For $I_1$, substitute $u=\tan(x/2)$ (Weierstrass substitution) and perform Partial Fraction Decomposition twice.

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For $I_2$, rewrite everything in terms of $\tan x$ and $\sec x$, let $v=\tan x$ and perform Partial Fraction Decomposition.

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The antiderivative is not very pleasing, have a look: $$I=\ln\left(\left|\sec\left(x\right)\right|\right)-\dfrac{\ln\left(\tan^2\left(x\right)+\tan\left(x\right)+1\right)}{2}+\dfrac{1}{\sqrt{3}}\left[\arctan\left(\frac{2\tan\left(x\right)+1}{\sqrt{3}}\right)-\ln\left(\left(\tan\left(\frac{x}{2}\right)-1\right)\left(\tan\left(\frac{x}{2}\right)+\sqrt{3}\right)+2\right)\\ +\ln\left(\left(\tan\left(\frac{x}{2}\right)-1\right)\left(\tan\left(\frac{x}{2}\right)-\sqrt{3}\right)+2\right)\right]+\ln\left(\left|\tan\left(\dfrac{x}{2}\right)+1\right|\right)-\ln\left(\left|\tan\left(\dfrac{x}{2}\right)-1\right|\right)$$

Answer 3

Hint Since the integrand is a rational function in the standard trigonometric functions applying the Weierstrass substitution, $$x = 2 \arctan t, \qquad dx = \frac{2}{1 + t^2} ,$$ rationalizes it: $$-16 \int \frac{t^2\, dt}{(t^4 - 2 t^3 + 2 t^2 + 2 t + 1) (t^2 + 1) (t - 1) (t + 1)} .$$ The quartic factor does not factor over the rationals, but it factors over the reals into a product of irreducible quadratic polynomials as $$t^4 - 2 t^3 + 2 t^2 + 2 t + 1 = [t^2 + (-1 + \sqrt{3}) t + (2 - \sqrt{3})][t^2 + (1 + \sqrt{3})t + (2 + \sqrt{3})].$$ With the denominator fully factored, we can now apply standard techniques: Decompose the integrand using the Method of Partial Fractions---which in this case involves solving for $8$ unknowns, and then integrate each term separately. In particular, all of the remaining computation is straightforward (but, in this case, pretty tedious).

Answer 4

Note that

$$\frac{\sin x}{\cos x+\csc x} = \frac{\sin^2x}{\sin x\cos x+1} = \frac1{2+\sin 2x}-\frac{\cos2x}{2+\sin 2x} $$

$$\frac{\tan x}{\cos x+\csc x} = \frac{\sec x(1-\cos^2x)}{\sin x\cos x+1} = \sec x-\frac{\sin x+ \cos x}{\sin x\cos x+1} $$ which results in four manageable integrals below

\begin{align} &\int\frac{\sin x+\tan x}{\cos x+\csc x}\ dx\\ =& \int \frac1{2+\sin 2x}-\frac{\cos2x}{2+\sin 2x} +\sec x-\frac{\sin x+ \cos x}{\sin x\cos x+1}\ dx\\ =& \ \frac1{\sqrt3}\tan^{-1}\frac{2\tan x+1}{\sqrt3} -\frac12\ln(2+\sin 2x)\\ &\ \>\>\> + \tanh^{-1}(\sin x)-\frac2{\sqrt3}\tanh^{-1}\frac{\sin x- \cos x}{\sqrt3} \end{align}