# Calculus/Integration techniques/Trigonometric Substitution/Solutions

1. ${\displaystyle \int {\frac {10}{25x^{2}+a}}dx}$
Let
${\displaystyle 5x={\sqrt {a}}\tan(\theta )\qquad 5dx={\sqrt {a}}\sec ^{2}(\theta )d\theta \qquad dx={\frac {{\sqrt {a}}\sec ^{2}(\theta )}{5}}d\theta }$

Then

{\displaystyle {\begin{aligned}\int {\frac {10}{25x^{2}+a}}dx&=10\int {\frac {{\sqrt {a}}\sec ^{2}(\theta )}{5(a\tan ^{2}(\theta )+a)}}d\theta \\&=2\int {\frac {{\sqrt {a}}\sec ^{2}(\theta )}{a\sec ^{2}(\theta )}}d\theta \\&={\frac {2{\sqrt {a}}}{a}}\int d\theta ={\frac {2\theta }{\sqrt {a}}}+C\\&\mathbf {={\frac {2}{\sqrt {a}}}\arctan {\Bigl (}{\frac {5x}{\sqrt {a}}}{\Bigr )}+C} \end{aligned}}}
Let
${\displaystyle 5x={\sqrt {a}}\tan(\theta )\qquad 5dx={\sqrt {a}}\sec ^{2}(\theta )d\theta \qquad dx={\frac {{\sqrt {a}}\sec ^{2}(\theta )}{5}}d\theta }$

Then

{\displaystyle {\begin{aligned}\int {\frac {10}{25x^{2}+a}}dx&=10\int {\frac {{\sqrt {a}}\sec ^{2}(\theta )}{5(a\tan ^{2}(\theta )+a)}}d\theta \\&=2\int {\frac {{\sqrt {a}}\sec ^{2}(\theta )}{a\sec ^{2}(\theta )}}d\theta \\&={\frac {2{\sqrt {a}}}{a}}\int d\theta ={\frac {2\theta }{\sqrt {a}}}+C\\&\mathbf {={\frac {2}{\sqrt {a}}}\arctan {\Bigl (}{\frac {5x}{\sqrt {a}}}{\Bigr )}+C} \end{aligned}}}