Indefinite Integrals
9.0 Integration of trigonometric functions
9.0 Integration of trigonometric functions
| S.No | Types of integral | Method to solve |
| 1. |
| Divide numerator and denominator by ${{{\cos }^2}x}$ and replace ${{{\sec }^2}x}$, if any, in denominator by $(1 + {\tan ^2}x)$ and substitute $\tan x = t \Rightarrow {\sec ^2}xdx = dt$ |
| 2. |
| Put $$\sin x = \frac{{2\tan \frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}$$ and $$\cos x = \frac{{1 - {{\tan }^2}\frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}$$ and substitute $\tan \frac{x}{2} = t$ to reduce the given integrals to $\int {\frac{1}{{a{t^2} + bt + c}}dt} $ which can be solved using the methods discussed earlier |
| 3. | $$\int {\frac{{a\cos x + b\sin x + c}}{{p\cos x + q\sin x + r}}dx} $$ | Write numerator = $\lambda $(denominator)+$\mu \frac{d}{{dx}}({\text{denominator)}}$+$\gamma $. Find $\lambda ,\mu ,\gamma $ by comparing the coefficients and solve the integral by splitting into sum or difference of three different integrals. |
| 4. | $$\int {\frac{{a\cos x + b\sin x}}{{p\cos x + q\sin x}}dx} $$ | Write numerator = $\lambda $(denominator)+$\mu \frac{d}{{dx}}({\text{denominator)}}$. Find $\lambda ,\mu $ by comparing the coefficients and solve the integral by splitting into sum or difference of two different integrals. |
Question 9. Evaluate $$\int {\frac{{3\cos x + 2}}{{\sin x + 2\cos x + 3}}dx} $$
Solution: To solve the integral, we can write it as $$3\cos x + 2 = \lambda (\sin x + 2\cos x + 3) + \mu (\cos x - 2\sin x) + \gamma $$
Comparing the coefficients, we get $$\lambda = \frac{6}{5},\mu = \frac{3}{5},\gamma = - \frac{8}{5}$$
$$\begin{equation} \begin{aligned} I = \int {\frac{{\lambda (\sin x + 2\cos x + 3) + \mu (\cos x - 2\sin x) + \gamma }}{{\sin x + 2\cos x + 3}}dx} \\ I = \lambda \int {dx} + \mu \int {\frac{{\cos x - 2\sin x}}{{\sin x + 2\cos x + 3}}dx} + \gamma \int {\frac{{dx}}{{\sin x + 2\cos x + 3}}} \\ I = \lambda x + \mu {I_1} + \gamma {I_2} \\\end{aligned} \end{equation} $$
Now, take $${I_1} = \int {\frac{{\cos x - 2\sin x}}{{\sin x + 2\cos x + 3}}dx} $$
Put $\sin x + 2\cos x + 3 = t \Rightarrow (\cos x - 2\sin x)dx = dt$, we get $${I_1} = \ln \left| {\sin x + 2\cos x + 3} \right|$$
Now take $${I_2} = \int {\frac{{dx}}{{\sin x + 2\cos x + 3}}} $$
Put $\sin x = \frac{{2\tan \frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}$ and $\cos x = \frac{{1 - {{\tan }^2}\frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}$, we get
$$\begin{equation} \begin{aligned} {I_2} = \int {\frac{{dx}}{{\frac{{2\tan \frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}} + 2.\frac{{1 - {{\tan }^2}\frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}} + 3}}} \\ {I_2} = \int {\frac{{{{\sec }^2}\frac{x}{2}}}{{{{\tan }^2}\frac{x}{2} + 2\tan \frac{x}{2} + 5}}dx} \\\end{aligned} \end{equation} $$ Now, put $\tan \frac{x}{2} = t \Rightarrow \frac{1}{2}{\sec ^2}\frac{x}{2} = dt$
$$\begin{equation} \begin{aligned} {I_2} = \int {\frac{{2dt}}{{{t^2} + 2t + 5}}} = 2\int {\frac{{dt}}{{{{(t + 1)}^2} + {2^2}}}} \\ {I_2} = \frac{2}{2}{\tan ^{ - 1}}\left( {\frac{{t + 1}}{2}} \right) = {\tan ^{ - 1}}\left( {\frac{{\tan \frac{x}{2} + 1}}{2}} \right) \\\end{aligned} \end{equation} $$
Therefore,
$$I = \lambda x + \mu \ln \left| {\sin x + 2\cos x + 3} \right| + \gamma {\tan ^{ - 1}}\left( {\frac{{\tan \frac{x}{2} + 1}}{2}} \right) + C$$
