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# 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 _____. 

Last updated date: 20th Jun 2024
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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. 

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 $\tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A}$ .