Question

Integrate:
\[\int{\dfrac{1}{3\sin x+4\cos x}}dx\]

Answer 1

Hint:
Use the half angle formula to simplify the expression.
Use the substitution method to solve the given integral.

Complete step by step answer:
Given that the integral is
\[\int{\dfrac{1}{3\sin x+4\cos x}}dx\]
In order to solve the given integral we need to use the following half angle formula followed the substitution method.
We know that
\[\cos x=\dfrac{1-{{\tan }^{2}}x/2}{1+{{\tan }^{2}}x/2}\]
\[\sin x=\dfrac{2\tan x/2}{1+{{\tan }^{2}}x/2}\]
Substitute the values in the given expression:
\[\int{\dfrac{dx}{3\left( \dfrac{2\tan x/2}{1+{{\tan }^{2}}x/2} \right)+4\left( \dfrac{1-{{\tan }^{2}}x/2}{1+{{\tan }^{2}}x/2} \right)}}\]
By simplifying the expression, we get
\[\int{\dfrac{\left( 1+{{\tan }^{2}}x/2 \right)\cdot dx}{4-4{{\tan }^{2}}x/2+6\tan \,x/2}}\]
Rewrite the expression by using the identity\[\left( 1+{{\tan }^{2}}A={{\sec }^{2}}A \right)\]:
\[\int{\dfrac{{{\sec }^{2}}x/2\,\,dx}{4-4{{\tan }^{2}}x/2+6\tan \,x/2}}\]
Let
\[\tan \,\,x/2=U\]
Differentiate both sides with respect to x
\[{{\sec }^{2}}x/2\,\dfrac{1}{2}dx=dU\]
Simplify the expression:
\[{{\sec }^{2}}x/2dx=2dU\]
Substitute these all values in the integral expression:
\[\int{\dfrac{2dU}{4-4{{U}^{2}}+6U}}\]
By taking 2 common from the denominator of expression, we get
\[\int{\dfrac{2dU}{2\left( 2-2{{U}^{2}}+3U \right)}}\]
Rewrite the integral after simplification:
\[\int{\dfrac{dU}{2-2{{U}^{2}}+3U}}\]
Rewrite the quadratic equation in the integral:
\[-\int{\dfrac{dU}{2{{U}^{2}}-3U-2}}\]
Solve the quadratic equation by splitting the middle term method:
\[-\int{\dfrac{dU}{2{{U}^{2}}-4U+U-2}}\]
\[\Rightarrow -\int{\dfrac{dU}{2U\left( U-2 \right)+1\left( U-2 \right)}}\]
\[\Rightarrow -\int{\dfrac{dU}{\left( 2U+1 \right)\left( U-2 \right)}}\]
Now, solve the expression \[\dfrac{1}{\left( 2U+1 \right)\left( U-2 \right)}\] by using partial fraction:
\[\dfrac{1}{\left( 2U+1 \right)\left( U-2 \right)}=\dfrac{A}{2U+1}+\dfrac{B}{U-2}\ldots \ldots \left( a \right)\]
Solve the equation for the value of \[A\]and \[B\]:
\[\dfrac{1}{\left( 2U+1 \right)\left( U-2 \right)}=\dfrac{A\left( U-2 \right)+B\left( 2U+1 \right)}{\left( 2U+1 \right)\left( U-2 \right)}\]
Again solve the equation:
\[\begin{align}
  & 1=A\left( U-2 \right)+B\left( 2U+1 \right) \\
 & \Rightarrow 1=AU-2A+2BU+B \\
 & \Rightarrow 1=\left( A+2B \right)U-2A+B \\
\end{align}\]
Compare the coefficients of both sides of the equation:
\[\begin{align}
  & A+2B=0\ldots \ldots \left( 1 \right) \\
 & -2A+B=1\ldots \ldots \left( 2 \right) \\
\end{align}\]
Multiply by 2 in (1) both sides followed the addition of both equations:
\[\begin{align}
  & 5B=1 \\
 & \Rightarrow B=\dfrac{1}{5} \\
\end{align}\]
Substitute the value \[B=\dfrac{1}{5}\]in equation (2), we get
\[\begin{align}
  & -2A+\dfrac{1}{5}=1 \\
 & \Rightarrow 2A=\dfrac{1}{5}-1 \\
 & \Rightarrow 2A=-\dfrac{4}{5} \\
 & \Rightarrow A=-\dfrac{2}{5} \\
\end{align}\]
Now, substitute the values of\[A\]and \[B\]in equation (a), we get
\[\dfrac{1}{\left( 2U+1 \right)\left( U-2 \right)}=\dfrac{-2}{5\left( 2U+1 \right)}+\dfrac{1}{5\left( U-2 \right)}\]
Now the expression of integral will be
\[-\int{\left\{ \dfrac{-2}{5\left( 2U+1 \right)}+\dfrac{1}{5\left( U-2 \right)} \right\}}dU\]
Break the integral expression in two parts:
\[\int{\dfrac{2dU}{5\left( 2U+1 \right)}}-\int{\dfrac{dU}{5\left( U-2 \right)}}\]
Rewrite the expression after simplification:
\[\dfrac{2}{5}\int{\dfrac{dU}{\left( 2U+1 \right)}}-\dfrac{1}{5}\int{\dfrac{dU}{\left( U-2 \right)}}\]
Solve the integral by using the standard formula:
\[\dfrac{2}{5}\dfrac{\ln \left( 2U+2 \right)}{2}-\dfrac{1}{5}\ln \left( U-2 \right)+C\]
Simplify the expression:
\[\dfrac{1}{5}\ln \left( 2U+2 \right)-\dfrac{1}{5}\ln \left( U-2 \right)+C\]
Simplify the expression:
\[\dfrac{1}{5}\left[ \ln \left( 2U+1 \right)-\ln \left( U-2 \right) \right]+C\]
Rewrite the expression by using logarithm property:
\[\dfrac{1}{5}\left[ \ln \left\{ \dfrac{2U+1}{U-2} \right\} \right]+C\]
Now, substitute the value of \[U\]in terms of \[x,\]we get
\[\dfrac{1}{5}\left[ \ln \left\{ \dfrac{2\tan x/2+1}{\tan x/2-2} \right\} \right]+C\]
Hence
\[\int{\dfrac{1}{3\sin x+4\cos x}}dx=\dfrac{1}{5}\left[ \ln \left[ \dfrac{2\tan x/2+1}{\tan x/2-1} \right]+C \right.\]

Note:
When a denominator of fraction in integral is given in trigonometric terms then we always use the half angle formula to convert integral in simple form.
This type of integral is always solved by substitution method.