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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
If $\int\frac{\sin^8\text{x}-\cos^8\text{x}}{1-2\sin^2\text{x}\cos^2\text{x}}\text{ dx}=\text{a}\sin2\text{x}+\text{C},$ then $a =$
  • ✓
    $-\frac{1}{2}$
  • B
    $\frac{1}{2}$
  • C
    $-1$
  • D
    $1$
✓

Answer

Correct option: A.
$-\frac{1}{2}$
$\int\frac{\sin^8\text{x}-\cos^8\text{x}}{1-2\sin^2\text{x}\cos^2\text{x}}\text{ dx}=\text{a}\sin2\text{x}+\text{C}\ ...(\text{i})$
Considering $\text{LHS}$ of $eq. (i)$
$\Rightarrow\int\frac{(\sin^4\text{x}-\cos^4\text{x})(\sin^4\text{x}+\cos^4\text{x})}{(1-2\sin^2\text{x}\cos^2\text{x})}$
$\Rightarrow\frac{(\sin^2\text{x}-\cos^2\text{x})(\sin^2\text{x}+\cos^2\text{x})\cdot(\sin^4\text{x}+\cos^4\text{x})\text{ dx}}{\big\{(\sin^2\text{x}+\cos^2\text{x})^2-2\sin^2\text{x}\cos^2\text{x}\big\}}$
$\Rightarrow\int\frac{(\sin^2\text{x}-\cos^2\text{x})\cdot(\sin^4\text{x}+\cos^4\text{x})\text{ dx}}{(\sin^4\text{x}+\cos^4\text{x}+2\sin^2\text{x}\cos^2\text{x}-2\sin^2\text{x}\cos^2\text{x})}$
$\Rightarrow-\int\frac{(\cos^2\text{x}-\sin^2\text{x})\times(\sin^4\text{x}+\cos^4\text{x})\text{ dx}}{(\sin^4\text{x}+\cos^4\text{x})}$
$\Rightarrow-\int\cos(2\text{x})\text{ dx}\ ...(\text{ii})$ $(\because\cos^2\text{x}-\sin^2\text{x}=\cos2\text{x})$
Comparing the $\text{RHS}$ of $eq. (i)$ with $eq. (ii)$ we get
$\text{a}=-\frac{1}{2}$

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