Here you will learn what is integration by substitution method class 12 with examples.
Let’s begin –
Integration By Substitution
The method of evaluating an integral by reducing it to standard form by a proper substitution is called integration by substitution.
If \(\phi(x)\) is continuously differentiable function, then to evaluate integrals of the form
\(\int\) \(f(\phi(x))\) \(\phi'(x)\) dx, we substitute \(\phi(x)\) = t and \(\phi'(x)\) dx = dt
This substitution reduces the above integral to \(\int\) f(t) dt.
After evaluating this integral we substitute back the value of t.
Also Read : Integration Formulas for Class 12 – Indefinite Integration
Example : Prove that \(\int\) sin(ax + b) dx = \(-1\over a\) cos(ax + b) + C.
Solution : Let ax + b = t. Then, d(ax + b) = dt \(\implies\) a dx = dt \(\implies\) dx = \(1\over a\) dt
Putting ax + b = t and dx = \(1\over a\) dt, we get
\(\int\) sin(ax + b) dx = \(1\over a\) \(\int\) sin t dt
= \(-1\over a\) cos t + C
= \(-1\over a\) cos(ax + b) + C
Example : Evaluate \(\int\) \(cos^2x\over {sin^2x + sinx}\) dx
Solution : I = \(\int\) \((1-sin^2x)cosx\over {sinx(1 + sinx)}\) dx = \(\int\) \(1 – sinx\over {sinx}\) cosx dx
Put sinx = t \(\implies\) cosx dx = dt
\(\implies\) I = \(\int\) \(1 – t\over t\) dt
= ln| t | – t + C
= ln|sinx| – sinx + C
