Some Special Integrals – Problems
\(\int{{{e}^{ax+b}}\left[ af(x)+f'(x) \right]}.dx={{e}^{ax+b}}f(x)+c\).
Example 1: \(\int{{{e}^{x}}\left[ \frac{1+\sin x}{1+\cos x} \right]dx}\) is equal to
Solution: Given \(\int{{{e}^{x}}\left[ \frac{1+\sin x}{1+\cos x} \right]dx}\).
\(\int{{{e}^{x}}\left[ \frac{1}{1+\cos x}+\frac{\sin x}{1+\cos x} \right]dx}\).
\(\int{{{e}^{x}}\left[ \frac{1}{2{{\cos }^{2}}\left( \frac{x}{2} \right)}+\frac{2\sin \left( \frac{x}{2} \right)\cos \left( \frac{x}{2} \right)}{2{{\cos }^{2}}\left( \frac{x}{2} \right)} \right]dx}\).
(cos2x = 2cos²x – 1 and sin2x = 2sinx cosx)
\(\int{{{e}^{x}}\left[ \frac{{{\sec }^{2}}\left( \frac{x}{2} \right)}{2}+\tan \left( \frac{x}{2} \right) \right]dx}\).
\(\int{{{e}^{x}}\left( \tan \left( \frac{x}{2} \right)+\frac{1}{2}{{\sec }^{2}}\left( \frac{x}{2} \right) \right)}dx\).
Let \(f(x)=\tan \left( \frac{x}{2} \right)\).
\(\Rightarrow f'(x)=\frac{{{\sec }^{2}}\left( \frac{x}{2} \right)}{2}\).
\(\therefore \int{{{e}^{x}}\left( \tan \left( \frac{x}{2} \right)+\frac{{{\sec }^{2}}\left( \frac{x}{2} \right)}{2} \right)}dx\).
\(=\frac{{{\sec }^{2}}\left( \frac{x}{2} \right)}{2}+c\).
\(\left( \because \int{{{e}^{x}}\left( f(x)+f'(x) \right)dx={{e}^{x}}f(x)+c} \right)\).
Example 2: \(\int{{{e}^{2x}}\sin xdx}\) is equal to
Solution: Given, \(\int{{{e}^{2x}}\sin xdx}\).
\(\int{{{e}^{2x}}\sin xdx}=\sin x\int{{{e}^{2x}}dx-}\int{(\frac{d}{dx}(\sin x).\int{{{e}^{2x}}dx)dx}}\).
\(=\sin x\frac{{{e}^{2x}}}{2}-\frac{1}{2}\int{(\frac{d}{dx}(\sin x).{{e}^{2x}})dx}\).
\(=\sin x\frac{{{e}^{2x}}}{2}-\frac{1}{2}\int{(\cos x.{{e}^{2x}})dx}\).
\(\int{(\sin x.{{e}^{2x}})dx}=\sin x.\frac{{{e}^{2x}}}{2}-\frac{1}{2}{{I}_{1}}\) … (1)
Where \({{I}_{1}}=\int{(\cos x.{{e}^{2x}})dx}\).
\({{I}_{1}}=\cos x\frac{{{e}^{2x}}}{2}-\int{(\frac{d}{dx}\cos x.\int{{{e}^{2x}}})dx}+{{c}_{1}}\).
\({{I}_{1}}=\cos x\frac{{{e}^{2x}}}{2}+\frac{1}{2}\int{\sin x{{e}^{2x}}dx}+{{c}_{1}}\) … (2)
Equation (2) substitute in equation (1)
\(\int{(\sin x.{{e}^{2x}})dx}=\sin x.\frac{{{e}^{2x}}}{2}-\frac{1}{2}(\cos x\frac{{{e}^{2x}}}{2}+\frac{1}{2}\int{\sin x{{e}^{2x}}dx})+{{c}_{1}}\).
\(=\sin x.\frac{{{e}^{2x}}}{2}-\frac{1}{4}\cos x\frac{{{e}^{2x}}}{2}-\frac{1}{4}\int{\sin x{{e}^{2x}}dx}-\frac{1}{2}{{c}_{1}}\).
\(\int{(\sin x.{{e}^{2x}})dx}+\frac{1}{4}\int{\sin x{{e}^{2x}}dx}==\sin x.\frac{{{e}^{2x}}}{2}-\frac{1}{4}\cos x{{e}^{2x}}-\frac{1}{2}{{c}_{1}}\).
\(\frac{5}{4}\int{\sin x{{e}^{2x}}dx}=\frac{\sin x\ {{e}^{2x}}}{2}-\frac{1}{4}\cos x{{e}^{2x}}-\frac{1}{2}{{c}_{1}}\).
\(\int{\sin x{{e}^{2x}}dx}=\frac{4}{5}\frac{\sin x\ {{e}^{2x}}}{2}-\frac{4}{20}\cos x{{e}^{2x}}-\frac{4}{10}{{c}_{1}}\).
\(\int{\sin x{{e}^{2x}}dx}=\frac{4}{5}\frac{\sin x\ {{e}^{2x}}}{2}-\frac{1}{5}\cos x{{e}^{2x}}-\frac{2}{5}{{c}_{1}}\).
\(\int{\sin x{{e}^{2x}}dx}=\frac{2}{5}\sin x\ {{e}^{2x}}-\frac{1}{5}\cos x{{e}^{2x}}+c\).