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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\sqrt {\frac{{\cos x - {{\cos }^3}x}}{{1 - {{\cos }^3}x}}} \;dx} $ is equal to
  • A
    $\frac{2}{3}{\sin ^{ - 1}}({\cos ^{3/2}}x) + c$
  • B
    $\frac{3}{2}{\sin ^{ - 1}}({\cos ^{3/2}}x) + c$
  • ✓
    $\frac{2}{3}{\cos ^{ - 1}}({\cos ^{3/2}}x) + c$
  • D
    None of these
✓

Answer

Correct option: C.
$\frac{2}{3}{\cos ^{ - 1}}({\cos ^{3/2}}x) + c$
c
(c)$\int_{}^{} {\sqrt {\frac{{\cos x - {{\cos }^3}x}}{{1 - {{\cos }^3}x}}} } \,dx = \int_{}^{} {\sqrt {\frac{{\cos x}}{{1 - {{\cos }^3}x}}} } \,\sin x\,dx$
$ = - \int_{}^{} {\sqrt {\frac{t}{{1 - {t^3}}}} } \,dt = - \int_{}^{} {\frac{{\sqrt t }}{{\sqrt {1 - {{({t^{3/2}})}^2}} }}\,dt} = - \frac{2}{3}\int_{}^{} {\frac{{\frac{3}{2}\sqrt t }}{{\sqrt {1 - {{({t^{3/2}})}^2}} }}\,dt} $
$ = \frac{2}{3}{\cos ^{ - 1}}({t^{3/2}}) + c = \frac{2}{3}{\cos ^{ - 1}}({\cos ^{3/2}}x) + c$.

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