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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^a \frac{x^4 d x}{\left( a ^2+x^2\right)^4}=$
  • ✓
    $\frac{1}{16 a ^3}\left(\frac{\pi}{4}-\frac{1}{3}\right)$
  • B
    $\frac{1}{16 a ^3}\left(\frac{\pi}{4}+\frac{1}{3}\right)$
  • C
    $\frac{1}{16} a^3\left(\frac{\pi}{4}-\frac{1}{3}\right)$
  • D
    $\frac{1}{16} a^3\left(\frac{\pi}{4}+\frac{1}{3}\right)$
✓

Answer

Correct option: A.
$\frac{1}{16 a ^3}\left(\frac{\pi}{4}-\frac{1}{3}\right)$
(A)
Put $x= a \tan \theta \Rightarrow d x= a \sec ^2 \theta d \theta$
$\therefore \quad \int_0^{ a } \frac{x^4}{\left( a ^2+x^2\right)^4} d x=\int_0^{\frac{\pi}{4}} \frac{ a ^4 \tan ^4 \theta \cdot a \sec ^2 \theta}{ a ^8 \sec ^8 \theta} d \theta$
$\begin{array}{l}=\frac{1}{ a ^3} \int_0^{\frac{\pi}{4}} \sin ^4 \theta \cos ^2 \theta d \theta \\ =\frac{1}{ a ^3} \int_0^{\frac{\pi}{4}}\left(\sin ^4 \theta-\sin ^6 \theta\right) d \theta\end{array}$
$=\frac{1}{a^3} \int_0^{\frac{\pi}{4}}\left[\frac{(1-\cos 2 \theta)^2}{4}-\frac{(1-\cos 2 \theta)^3}{8}\right] d \theta$
$\begin{array}{l}=\frac{1}{8 a ^3} \int_0^{\frac{\pi}{4}}(1+\cos 2 \theta)(1-\cos 2 \theta)^2 d \theta \\ =\frac{1}{8 a ^3} \int_0^{\frac{\pi}{4}}\left(1-\cos 2 \theta+\cos ^2 2 \theta+\cos ^3 2 \theta\right) d \theta\end{array}$
$=\frac{1}{8 a ^3} \int_0^{\frac{\pi}{4}} \frac{1}{4}[2-\cos 2 \theta-2 \cos 4 \theta+\cos 6 \theta] d \theta$
$\ldots\left[\begin{array}{c}\because \cos ^2 A=\frac{1+\cos 2 A}{2} \\ \text { and } \cos ^2 A=\frac{\cos 3 A+3 \cos A}{4}\end{array}\right]$
$\begin{array}{l}=\frac{1}{32 a ^3}\left[2 \theta-\frac{\sin 2 \theta}{2}-\frac{\sin 4 \theta}{2}+\frac{\sin 6 \theta}{6}\right]_0^{\frac{\pi}{4}} \\ =\frac{1}{16 a ^3}\left(\frac{\pi}{4}-\frac{1}{3}\right)\end{array}$

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