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Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 4 Maths Textbook Solution.

Answers (1)


A=\frac{2}{3}, B=\frac{5}{3}


\int \frac{1}{5+4 \sin x} d x=A \tan ^{-1}\left(B \tan \frac{x}{2}+\frac{4}{3}\right)+C


Using \int \frac{1}{1+t^{2}} d t


Let \mathrm{I}=\int \frac{1}{5+4 \sin x} d x

         \begin{aligned} &=\int \frac{1}{5+4.2 \sin \frac{x}{2} \cos \frac{x}{2}} d x \\ &=\int \frac{1}{5.1+8 \sin \frac{x}{2} \cos \frac{x}{2}} d x \end{aligned}                                                    [\because \sin 2 \theta=2 \sin \theta \cos \theta]

         =\int \frac{1}{5\left(\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x^{2}}{2}\right)+8 \sin \frac{x}{2} \cos _{2}^{x}} d x               \left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]

        =\int \frac{1}{5 \sin ^{2} \frac{x}{2}+5 \cos ^{2} \frac{x}{2}+8 \sin \frac{x}{2} \cos _{2}^{x}} d x

Divide num. and den. by \cos ^{2} \frac{x}{2}

      \begin{aligned} &=\int \frac{\frac{1}{\cos ^{2} \frac{x}{2}}}{5 \tan ^{2} \frac{x}{2}+5+8 \tan _{2}^{x}} d x \\ &=\frac{1}{5} \int \frac{\sec ^{2} \frac{x}{2}}{\tan ^{2} \frac{x}{2}+\frac{8}{5} \tan _{2}^{x}+1} d x \end{aligned}             \left[\because \sec x=\frac{1}{\cos x}\right]

     =\frac{1}{5} \int \frac{2}{t^{2}+\frac{8}{5} t+1} d t                                                \left[\text { Put } \tan \frac{x}{2}=t \Rightarrow \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t \Rightarrow \sec ^{2} \frac{x}{2} d x=2 d t\right]

      \begin{aligned} &=\frac{2}{5} \int \frac{d t}{t^{2}+\frac{8 t}{5}+\left(\frac{4}{5}\right)^{2}-\left(\frac{4}{5}\right)^{2}+1} \\ &=\frac{2}{5} \int \frac{d t}{\left(t+\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}} \\ &=\frac{2}{5} \times \frac{1}{\frac{3}{5}} \tan ^{-1}\left(\frac{5 t+4}{3}\right)+C \quad \ldots \ldots \ldots \ldots \ldots\left\{\frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c\right\} \end{aligned}

\therefore I=\frac{2}{3} \tan ^{-1}\left(\frac{5}{3} \tan \frac{x}{2}+\frac{4}{3}\right)+C                         [E q \cdot(i)]

Acc. to given,

\mathrm{I}=A \tan ^{-1}\left(B \tan \frac{x}{2}+\frac{4}{3}\right)+C                             [E q \cdot(i i)]

From Eq. ( i ) and ( ii )

A \tan ^{-1}\left(B \tan \frac{x}{2}+\frac{4}{3}\right)+C=\frac{2}{3} \tan ^{-1}\left(\frac{5}{3} \tan \frac{x}{2}+\frac{4}{3}\right)+C

Comparing both sides, we get

A=\frac{2}{3}, B=\frac{5}{3}


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