Question

Let $ F\left( x \right) $ be the anti-derivative of $ f\left( x \right)=\dfrac{1}{3+5\sin x+3\cos x}dx $ whose graph passes through the point $ \left( 0,0 \right) $ . Then $ F\left( \dfrac{\pi }{2} \right)-\dfrac{1}{5}\log \dfrac{8}{3}+1982 $ is equal to _____. \[\]

Answer 1

Hint: We integrate $ f\left( x \right) $ with respect to $ x $ by substituting $ u=\tan \dfrac{x}{2} $ and find the anti-derivative $ F\left( x \right) $ with integration constant $ c $ .We find $ c $ using satisfaction of the given point $ \left( 0,0 \right) $ in the graph of $ F\left( x \right) $ . We put $ x=\dfrac{\pi }{2} $ and simplify to get the result. \[\]


Complete step by step answer:
We know that anti-derivative, primitive function or indefinite integral of a function $ f $ is a differentiable function $ F $ whose derivative is equal to the original function $ f $ which means $ {{F}^{'}}=f $ . The process of finding integral is called integration and the original function $ f $ is called integrand. We write integration with respect to variable $ x $ as
\[\int{f\left( x \right)}dx=F\left( x \right)+c\]
We know from double angle formula of sine and cosine in terms tangent for some angle $ A $ as;
\[\begin{align}
  & \sin 2A=\dfrac{2\tan A}{1+{{\tan }^{2}}A} \\
 & \cos 2A=\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A} \\
\end{align}\]
We are given the following original function
 $ f\left( x \right)=\dfrac{1}{3+5\sin x+3\cos x} $
Let us integrate the above function by u-substitution method to find the anti-derivative $ F\left( x \right) $ .
 $ F\left( x \right)=\int{f\left( x \right)dx}=\int{\dfrac{1}{3+5\sin x+3\cos x}dx} $
Let us have $ u=\tan \dfrac{x}{2} $ . We differentiate both sides with respect to $ x $ to have;
\[\begin{align}
  & \dfrac{d}{dx}u=\dfrac{d}{dx}\tan \dfrac{x}{2} \\
 & \Rightarrow \dfrac{du}{dx}=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2} \\
 & \Rightarrow \dfrac{du}{\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}}=dx \\
\end{align}\]
We use the Pythagorean trigonometric identity $ {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ to have;
\[\begin{align}
  & \Rightarrow \dfrac{du}{\dfrac{1}{2}\left( 1+{{\tan }^{2}}\dfrac{x}{2} \right)}=dx \\
 & \Rightarrow dx=\dfrac{2du}{1+{{u}^{2}}} \\
\end{align}\]
We use the double angle formula of sine and cosine in terms tangent for $ A=\dfrac{x}{2} $ to have;
 $ \begin{align}
  & \sin x=\dfrac{2\tan \dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}=\dfrac{2u}{1+{{u}^{2}}} \\
 & \cos x=\dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}=\dfrac{1-{{u}^{2}}}{1+{{u}^{2}}} \\
\end{align} $
We put $ \sin x,\cos x,dx $ in terms of $ u,du $ in the integrand to have

\[\begin{align}
  & F\left( x \right)=\int{\dfrac{1}{3+5\left( \dfrac{2u}{1+{{u}^{2}}} \right)+3\left( \dfrac{1-{{u}^{2}}}{1+{{u}^{2}}} \right)}\times \dfrac{2du}{1+{{u}^{2}}}} \\
 & \Rightarrow F\left( x \right)=\int{\dfrac{2du}{\dfrac{3+3{{u}^{2}}+10u+3-2\mathsf{}{{u}^{2}}}{1+{{u}^{2}}}\times \left( 1+{{u}^{2}} \right)}} \\
 & \Rightarrow F\left( x \right)=\int{\dfrac{2du}{6+10u}} \\
 & \Rightarrow F\left( x \right)=\int{\dfrac{du}{3+5u}} \\
\end{align}\]
We take 5 multiply in the numerator and the denominator of the integrand and have;
\[\begin{align}
  & \Rightarrow F\left( x \right)=\dfrac{1}{5}\int{\dfrac{5du}{\left( 5u+3 \right)}} \\
 & \Rightarrow F\left( x \right)=\dfrac{1}{5}\log \left( 5u+3 \right)+c \\
 & \Rightarrow F\left( x \right)=\dfrac{1}{5}\log \left( 5\tan \dfrac{x}{2}+3 \right)+c \\
\end{align}\]
Here $ c $ is an arbitrary integration constant. We are given that graph of $ F\left( x \right) $ passes through $ \left( 0,0 \right) $ . So we have
\[\begin{align}
  & F\left( 0 \right)=0 \\
 & \Rightarrow \dfrac{1}{5}\log \left( 5\tan \dfrac{0}{2}+3 \right)+c=0 \\
 & \Rightarrow c=\dfrac{-\log 3}{5} \\
\end{align}\]
 So we have the anti-derivative function as
\[F\left( x \right)=\dfrac{1}{5}\log \left( 5\tan \dfrac{x}{2}+3 \right)-\dfrac{\log 3}{5}\]
We put $ x=\dfrac{\pi }{2} $ in $ F\left( x \right) $ to have;
\[\begin{align}
  & F\left( \dfrac{\pi }{2} \right)=\dfrac{1}{5}\log \left( 5\tan \dfrac{\dfrac{\pi }{2}}{2}+3 \right)-\dfrac{\log 3}{5} \\
 & \Rightarrow F\left( \dfrac{\pi }{2} \right)=\dfrac{1}{5}\log \left( 5\tan \dfrac{\pi }{4}+3 \right)-\dfrac{\log 3}{5} \\
 & \Rightarrow F\left( \dfrac{\pi }{2} \right)=\dfrac{1}{5}\log \left( 5\cdot 1+3 \right)-\dfrac{\log 3}{5} \\
 & \Rightarrow F\left( \dfrac{\pi }{2} \right)=\dfrac{1}{5}\left( \log 8-\log 3 \right) \\
\end{align}\]
We use logarithmic identity of quotient $ \log \left( \dfrac{a}{b} \right)=\log a-\log b $ in the above step to have;
\[\begin{align}
  & \Rightarrow F\left( \dfrac{\pi }{2} \right)=\dfrac{1}{5}\log \left( \dfrac{8}{3} \right) \\
 & \Rightarrow F\left( \dfrac{\pi }{2} \right)-\dfrac{1}{5}\log \dfrac{8}{3}=0 \\
\end{align}\]
We add 1982 both sides of the above step to have;
\[\Rightarrow F\left( \dfrac{\pi }{2} \right)-\dfrac{1}{5}\log \dfrac{8}{3}+1982=1982\]
So the answer is 1982. \[\]

Note:
 We should remember the integral identity $ \int{\dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}dx=\log \left| f\left( x \right) \right|}+c $ . We note that integral remains same if we change the variable. The derivative of the function geometrically represents the slope of the tangent at any point; the anti-derivative represents the area under the curve. We must be careful of the confusion between sine double angle formula and tangent double angle formula which is given by $ \tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A} $ .