Exercise$\int\frac{1}{4sec\left(x\right)-1}dx$
Step-by-step Solution
1
We can solve the integral $\int\frac{1}{4\sec\left(x\right)-1}dx$ by applying the method Weierstrass substitution (also known as tangent half-angle substitution) which converts an integral of trigonometric functions into a rational function of $t$ by setting the substitution
$t=\tan\left(\frac{x}{2}\right)$
$\sin x=\frac{2t}{1+t^{2}},\:\cos x=\frac{1-t^{2}}{1+t^{2}},\:\mathrm{and}\:\:dx=\frac{2}{1+t^{2}}dt$
3
Substituting in the original integral we get
$\int\frac{1}{4\left(\frac{1+t^{2}}{1-t^{2}}\right)-1}\frac{2}{1+t^{2}}dt$
Intermediate steps
$\int\frac{2\left(1-t^{2}\right)}{\left(4\left(1+t^{2}\right)-\left(1-t^{2}\right)\right)\left(1+t^{2}\right)}dt$
Explain this step further
5
Take out the constant $2$ from the integral
$2\int\frac{1-t^{2}}{\left(4\left(1+t^{2}\right)-\left(1-t^{2}\right)\right)\left(1+t^{2}\right)}dt$
6
Solve the product $4\left(1+t^{2}\right)$
$2\int\frac{1-t^{2}}{\left(4+4t^{2}-\left(1-t^{2}\right)\right)\left(1+t^{2}\right)}dt$
7
Solve the product $-\left(1-t^{2}\right)$
$2\int\frac{1-t^{2}}{\left(4+4t^{2}-1+t^{2}\right)\left(1+t^{2}\right)}dt$
Intermediate steps
8
Simplify the expression
$2\int\frac{1-t^{2}}{\left(3+5t^{2}\right)\left(1+t^{2}\right)}dt$
Explain this step further
Intermediate steps
9
Rewrite the fraction $\frac{1-t^{2}}{\left(3+5t^{2}\right)\left(1+t^{2}\right)}$ in $2$ simpler fractions using partial fraction decomposition
$\frac{4}{3+5t^{2}}+\frac{-1}{1+t^{2}}$
Explain this step further
10
Expand the integral $\int\left(\frac{4}{3+5t^{2}}+\frac{-1}{1+t^{2}}\right)dt$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately
$2\int\frac{4}{3+5t^{2}}dt+2\int\frac{-1}{1+t^{2}}dt$
Intermediate steps
11
The integral $2\int\frac{4}{3+5t^{2}}dt$ results in: $\frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}t\right)}{3}$
$\frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}t\right)}{3}$
Explain this step further
Intermediate steps
12
The integral $2\int\frac{-1}{1+t^{2}}dt$ results in: $-2\arctan\left(t\right)$
$-2\arctan\left(t\right)$
Explain this step further
13
Gather the results of all integrals
$\frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}t\right)}{3}-2\arctan\left(t\right)$
14
Replace $t$ with the value that we assigned to it in the beginning: $\tan\left(\frac{x}{2}\right)$
$\frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}\tan\left(\frac{x}{2}\right)\right)}{3}-2\arctan\left(\tan\left(\frac{x}{2}\right)\right)$
Intermediate steps
15
Simplify the expression
$\frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}\tan\left(\frac{x}{2}\right)\right)}{3}-x$
Explain this step further
16
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}\tan\left(\frac{x}{2}\right)\right)}{3}-x+C_0$
Final answer to the exercise $\frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}\tan\left(\frac{x}{2}\right)\right)}{3}-x+C_0$