Question

Evaluate the given integral: $\int{\dfrac{dx}{3\cos x+4\sin x+6}}$

Answer 1

Hint: Start by letting $\tan \dfrac{x}{2}$ be t and solve the integral. Use the formula $\sin x=\dfrac{2\tan \dfrac{x}{2}}{{{\sec }^{2}}\dfrac{x}{2}}$ and $\cos x=\dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{{{\sec }^{2}}\dfrac{x}{2}}$ to change the integral in terms of t. After that use the formula $\dfrac{dx}{{{a}^{2}}+{{\left( x+k \right)}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}\dfrac{x+k}{a}+c$ .

Complete step-by-step answer:
Let us start with the integral given in the above question.
$\int{\dfrac{dx}{3\cos x+4\sin x+6}}$
We know that $\sin x=\dfrac{2\tan \dfrac{x}{2}}{{{\sec }^{2}}\dfrac{x}{2}}$ and $\cos x=\dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{{{\sec }^{2}}\dfrac{x}{2}}$ . So, using this in our integral, we get
$=\int{\dfrac{dx}{3\left( \dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{{{\sec }^{2}}\dfrac{x}{2}} \right)+4\left( \dfrac{2\tan \dfrac{x}{2}}{{{\sec }^{2}}\dfrac{x}{2}} \right)+6}}$
$=\int{\dfrac{{{\sec }^{2}}\dfrac{x}{2}dx}{3-3{{\tan }^{2}}\dfrac{x}{2}+8\tan \dfrac{x}{2}+6{{\sec }^{2}}\dfrac{x}{2}}}$
Now we will use the identity $1+{{\tan }^{2}}\alpha ={{\sec }^{2}}\alpha $ .
$=\int{\dfrac{{{\sec }^{2}}\dfrac{x}{2}dx}{3-3{{\tan }^{2}}\dfrac{x}{2}+8\tan \dfrac{x}{2}+6+6{{\tan }^{2}}\dfrac{x}{2}}}$
$=\int{\dfrac{{{\sec }^{2}}\dfrac{x}{2}dx}{3{{\tan }^{2}}\dfrac{x}{2}+8\tan \dfrac{x}{2}+9}}$
Now to convert the integral to a form from where we can directly integrate it, we let $\tan \dfrac{x}{2}$ to be t.
 $\tan \dfrac{x}{2}=t.......(i)$
Now if we differentiate both the sides of the equation, we get
$\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}dx=dt$
So, if we substitute the terms in the integral in terms of t, our integral becomes:
$\int{\dfrac{2dt}{3{{t}^{2}}+8t+9}}$
$=\int{\dfrac{2dt}{3\left( {{t}^{2}}+\dfrac{8}{3}t+3 \right)}}$
The denominator can be written as: ${{t}^{2}}+\dfrac{8}{3}t+3={{\left( t+\dfrac{4}{3} \right)}^{2}}+{{\left( \sqrt{\dfrac{11}{9}} \right)}^{2}}$ .
$\therefore \int{\dfrac{2dt}{3\left( {{t}^{2}}+\dfrac{8}{3}t+3 \right)}}=\int{\dfrac{2dt}{3\left( {{\left( t+\dfrac{4}{3} \right)}^{2}}+{{\left( \sqrt{\dfrac{11}{9}} \right)}^{2}} \right)}}$
Now we know that $\dfrac{dx}{{{a}^{2}}+{{\left( x+k \right)}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}\dfrac{x+k}{a}+c$ . So, our integral comes out to be:
 $\int{\dfrac{2dt}{3\left( {{\left( t+\dfrac{4}{3} \right)}^{2}}+{{\left( \dfrac{\sqrt{11}}{3} \right)}^{2}} \right)}}=\dfrac{2}{3}\left( \dfrac{3}{\sqrt{11}}{{\tan }^{-1}}\dfrac{t+\dfrac{4}{3}}{\dfrac{\sqrt{11}}{3}}+c \right)$
$\Rightarrow \int{\dfrac{2dt}{3\left( {{\left( t+\dfrac{4}{3} \right)}^{2}}+{{\left( \dfrac{\sqrt{11}}{3} \right)}^{2}} \right)}}=\dfrac{2}{\sqrt{11}}{{\tan }^{-1}}\left( \dfrac{4+3t}{\sqrt{11}} \right)+\dfrac{2}{3}c$
Now we will substitute the value of t as assumed by us to convert our answer in terms of x.
$\dfrac{2}{\sqrt{11}}{{\tan }^{-1}}\left( \dfrac{4+3t}{\sqrt{11}} \right)+\dfrac{2}{3}c$
As $\dfrac{2}{3}c$ is also constant, so we take it to be ${{c}_{1}}$ .
$=\dfrac{2}{\sqrt{11}}{{\tan }^{-1}}\left( \dfrac{4+3\tan \dfrac{x}{2}}{\sqrt{11}} \right)+{{c}_{1}}$
So, we can say that the value of $\int{\dfrac{dx}{3\cos x+4\sin x+6}}$ is equal to $=\dfrac{2}{\sqrt{11}}{{\tan }^{-1}}\left( \dfrac{4+3\tan \dfrac{x}{2}}{\sqrt{11}} \right)+{{c}_{1}}$ .

Note: Don’t forget to substitute the assumed variable in your integrated expression to reach the final answer. Also, you need to remember all the basic formulas that we use for indefinite integrals as they are used in definite integrations as well.