Question

Integration by Partial Fractions

1.) ∫dt/(1+sint+cost)

2.) ∫dy/(5+4cosy)

Answer 1

$1) \displaystyle\int\frac{dt}{1+\sin{t}+\cos{t}} \\ 2) \displaystyle\int\frac{dy}{5+4\cos{y}}$

Solution:

$1) \displaystyle\int\frac{dt}{1+\sin{t}+\cos{t}}$

Let's apply the change of variables:

$\tan{\frac t2}=x$; $\displaystyle\frac{dt}{2\cos^2{\frac t2}}=dx$ ; $\displaystyle\frac12(1+\tan^2{\frac t2})dt=dx$ ; $\displaystyle dt=\frac{2dx}{1+x^2}$ .

$\displaystyle\sin{t}=\frac{2\tan{\frac t2}}{1+\tan^2{\frac t2}}=\frac{2x}{1+x^2}$ ; $\displaystyle\cos{t}=\frac{1-\tan^2{\frac t2}}{1+\tan^2{\frac t2}}=\frac{1-x^2}{1+x^2}$ .

$\displaystyle\int\frac{dt}{1+\sin{t}+\cos{t}}=$ $\displaystyle\int\frac{2dx}{(1+x^2)(1+\frac{2x}{1+x^2}+\frac{1-x^2}{1+x^2})}=$ $\displaystyle\int\frac{2dx}{(2+2x)}=$ $\displaystyle\int\frac{dx}{(1+x)}=\ln|1+x|+C=\ln{\left|1+\tan{\frac t2}\right|}+C$ .

(C is constant).

$2) \displaystyle\int\frac{dy}{5+4\cos{y}}$

Let's apply the change of variables:

$\tan{\frac y2}=x$; $\displaystyle\frac{dy}{2\cos^2{\frac y2}}=dx$ ; $\displaystyle\frac12(1+\tan^2{\frac y2})dy=dx$ ; $\displaystyle dy=\frac{2dx}{1+x^2}$ .

$\displaystyle\cos{y}=\frac{1-\tan^2{\frac y2}}{1+\tan^2{\frac y2}}=\frac{1-x^2}{1+x^2}$ .

$\displaystyle\int\frac{dy}{5+4\cos{y}}=\int\frac{2dx}{(1+x^2)(5+4\cdot\frac{1-x^2}{1+x^2})}=$ $\displaystyle\int\frac{2dx}{9+x^2}=\frac23\int\frac{\frac{dx}{3}}{1+\frac{x^2}{9}}=$ $\displaystyle\frac23\arctan{\frac{x}{3}}+C=\frac23\arctan{\left(\frac{1}{3}\tan{\frac{y}{2}}\right)}+C$ .

(C is constant).

Answer:

1)$\displaystyle\int\frac{dt}{1+\sin{t}+\cos{t}}=\ln{\left|1+\tan{\frac t2}\right|}+C$ ;

$2) \displaystyle\int\frac{dy}{5+4\cos{y}}=\frac23\arctan{\left(\frac{1}{3}\tan{\frac{y}{2}}\right)}+C$ .