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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
$\int {{{\cos }^{ - 3/7}}} x{\sin ^{ - 11/7}}x\,\,dx = $
  • A
    $\log |{\sin ^{4/7}}x| + c$
  • B
    $\frac{{ - 7}}{4}{\cot ^{ - 4/7}}x + c$
  • ✓
    $\frac{{ - 7}}{4}{\tan ^{ - 4/7}}x + c$
  • D
    (b) and (c) both
✓

Answer

Correct option: C.
$\frac{{ - 7}}{4}{\tan ^{ - 4/7}}x + c$
c
(c) $m + n = - \frac{3}{7} + \left( {\frac{{ - 11}}{7}} \right) = - 2$ (ûve integer)
$I = \int {{{\cos }^{ - 3/7}}x\left( {{{\sin }^{( - 2 + 3/7)}}x} \right)dx} = \int {{{\cos }^{ - 3/7}}} x\,{\sin ^{ - 2}}x\,{\sin ^{3/7}}xdx$
$ = \int {\frac{{\cos e{c^2}x}}{{\left( {\frac{{{{\cos }^{3/7}}x}}{{{{\sin }^{3/7}}x}}} \right)}}} dx = \int {\frac{{\cos e{c^2}x\,dx}}{{{{\cot }^{3/7}}x}}} $
Put $\cot x = t$==> $ - \cos e{c^2}xdx = dt$
$I = - \int {\frac{{dt}}{{{t^{3/7}}}}} $$ = - \frac{{{t^{ - \frac{3}{7} + 1}}}}{{ - \frac{3}{7} + 1}} + c$$ = - \frac{7}{4}{t^{4/7}} + c$
$ = - \frac{7}{4}{\cot ^{4/7}}x + c$$ = - \frac{7}{4}{\tan ^{ - 4/7}}x + c$.
 

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