Integration by trigonometric substitution
Integration by trigonometric substitution is another method of solving problems involving integration,in my previous discussion i treated integration using euler substitution.
For any given function of the forms below:
\begin{align*}
&\sqrt{a^2-x^2},\ \text{use}\ x=a\sin\theta\\
&\sqrt{a^2+x^2},\ \text{use}\ x=a\tan\theta\\
&\sqrt{x^2-a^2},\ \text{use}\ x=a\sec\theta\\
\end{align*}
We use integration by trigonometric substitution to evaluate the integral.
Example
$\int\frac{dx}{(x^2+9)^\frac{3}{2}}$
Solution
$\int\frac{dx}{(x^2+9)^\frac{3}{2}}$.............(1)
we can see that this function is of the form, $\sqrt{a^2+x^2}$, so we use $x=a\tan\theta$
$\int\frac{dx}{(x^2+a^2)^\frac{3}{2}}=\int\frac{dx}{(x^2+3^2)^\frac{3}{2}}$
Let a=3, so that $x=3\tan\theta$...........(2)
and $dx=3\sec^2\theta{d\theta}$...........(3)
Now,lets separate the denominator of the integral before we proceed to carrying out integration. substitute $x=3\tan\theta$
\begin{align*}
&(x^2+9)=((3\tan\theta)^2+9)^\frac{3}{2}\\
&=(9\tan^2\theta+9)^\frac{3}{2}\\
&=(9[\tan^2\theta+1])^\frac{3}{2}\\
&=9^\frac{3}{2}[\sec^2\theta]^\frac{3}{2}\\
&=27\sec^3\theta\\
&\therefore (x^2+9)^\frac{3}{2}=27\sec^3\theta.........(4)\\
\text{subst. (3) and (4) into (1)}\\
&\int\frac{dx}{(x^2+9)^\frac{3}{2}}=\int\frac{3\sec^2\theta{d\theta}}{27\sec^3\theta}\\
&\int\frac{\sec^2\theta{d\theta}}{9\sec^3\theta}=\frac{1}{9}\int\frac{\sec^2\theta}{\sec^3\theta}d\theta\\
&=\frac{1}{9}\int\frac{1}{\sec\theta}d\theta=\frac{1}{9}\int\cos\theta{d\theta}\\
&\frac{1}{9}\sin\theta+c\\
\end{align*}
where c is constant of integration.
Now we bring back x, because our answer needs to be obtained in terms of x.
since we know that $x=3\tan\theta$.Let $\tan\theta=\frac{x}{3}$, now we know from our knowledge of trigonometry that $\tan\theta=\frac{x}{3}$,
so we can draw an angle to find the value of $\sin\theta$
Note that the hypotenuse was obtained using the Pythagorean identity. Hence by the rule of SOHCAHTOA $\sin\theta=\frac{x}{\sqrt{x^2+9}}$.
$\therefore$ we can conclude that \begin{align*}
&\int\frac{dx}{(x^2+9)^\frac{3}{2}}=\frac{1}{9}\sin\theta+c\\
&=\frac{1}{9}(\frac{x}{\sqrt{x^2+9}})+c=\frac{x}{9\sqrt{x^2+9}}+c
\end{align*}
Integration by trigonometric substitution is another method of solving problems involving integration,in my previous discussion i treated integration using euler substitution.
For any given function of the forms below:
\begin{align*}
&\sqrt{a^2-x^2},\ \text{use}\ x=a\sin\theta\\
&\sqrt{a^2+x^2},\ \text{use}\ x=a\tan\theta\\
&\sqrt{x^2-a^2},\ \text{use}\ x=a\sec\theta\\
\end{align*}
We use integration by trigonometric substitution to evaluate the integral.
Example
$\int\frac{dx}{(x^2+9)^\frac{3}{2}}$
Solution
$\int\frac{dx}{(x^2+9)^\frac{3}{2}}$.............(1)
we can see that this function is of the form, $\sqrt{a^2+x^2}$, so we use $x=a\tan\theta$
$\int\frac{dx}{(x^2+a^2)^\frac{3}{2}}=\int\frac{dx}{(x^2+3^2)^\frac{3}{2}}$
Let a=3, so that $x=3\tan\theta$...........(2)
and $dx=3\sec^2\theta{d\theta}$...........(3)
Now,lets separate the denominator of the integral before we proceed to carrying out integration. substitute $x=3\tan\theta$
\begin{align*}
&(x^2+9)=((3\tan\theta)^2+9)^\frac{3}{2}\\
&=(9\tan^2\theta+9)^\frac{3}{2}\\
&=(9[\tan^2\theta+1])^\frac{3}{2}\\
&=9^\frac{3}{2}[\sec^2\theta]^\frac{3}{2}\\
&=27\sec^3\theta\\
&\therefore (x^2+9)^\frac{3}{2}=27\sec^3\theta.........(4)\\
\text{subst. (3) and (4) into (1)}\\
&\int\frac{dx}{(x^2+9)^\frac{3}{2}}=\int\frac{3\sec^2\theta{d\theta}}{27\sec^3\theta}\\
&\int\frac{\sec^2\theta{d\theta}}{9\sec^3\theta}=\frac{1}{9}\int\frac{\sec^2\theta}{\sec^3\theta}d\theta\\
&=\frac{1}{9}\int\frac{1}{\sec\theta}d\theta=\frac{1}{9}\int\cos\theta{d\theta}\\
&\frac{1}{9}\sin\theta+c\\
\end{align*}
where c is constant of integration.
Now we bring back x, because our answer needs to be obtained in terms of x.
since we know that $x=3\tan\theta$.Let $\tan\theta=\frac{x}{3}$, now we know from our knowledge of trigonometry that $\tan\theta=\frac{x}{3}$,
so we can draw an angle to find the value of $\sin\theta$
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Note that the hypotenuse was obtained using the Pythagorean identity. Hence by the rule of SOHCAHTOA $\sin\theta=\frac{x}{\sqrt{x^2+9}}$.
$\therefore$ we can conclude that \begin{align*}
&\int\frac{dx}{(x^2+9)^\frac{3}{2}}=\frac{1}{9}\sin\theta+c\\
&=\frac{1}{9}(\frac{x}{\sqrt{x^2+9}})+c=\frac{x}{9\sqrt{x^2+9}}+c
\end{align*}
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