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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
$\int {\frac{{{{\sin }^8}\,x - {{\cos }^8}\,x}}{{\left( {1 - 2\,{{\sin }^2}\,x\,{{\cos }^2}\,x} \right)}}} dx$ is equal to
  • A
    $\frac{1}{2}\,\sin \,2x + c$
  • ✓
    $-\frac{1}{2}\,\sin \,2x + c$
  • C
    $-\frac{1}{2}\,\sin \,x + c$
  • D
    $ - \sin {^2}\,x + c$
✓

Answer

Correct option: B.
$-\frac{1}{2}\,\sin \,2x + c$
b
${\rm{ Let I}} = \int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$

$ = \int {\frac{{{{\left( {{{\sin }^4}x} \right)}^2} - {{\left( {{{\cos }^4}x} \right)}^2}}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} .dx$

$ = \int {\frac{{\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^4}x - {{\cos }^4}x} \right)}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} .dx$

$\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x{{\cos }^2}x} \right]$

$ = \int {\frac{{\left[ {\left( {{{\sin }^2}x + {{\cos }^2}x} \right]\left[ {{{\sin }^2}x - {{\cos }^2}x} \right]} \right.}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$

$ =  - \int {\cos } 2xdx = \frac{{ - \sin 2x}}{2} + c$

$ =  - \frac{1}{2}\sin 2x + c$

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