Integral Calculus Question 355
Question: $ \int_{{}}^{{}}{{{\sin }^{-1}}(3x-4x^{3})dx=} $
[AISSE 1986; DSSE 1984]
Options:
A) $ x{{\sin }^{-1}}x+\sqrt{1-x^{2}}+c $
B) $ x{{\sin }^{-1}}x-\sqrt{1-x^{2}}+c $
C) $ 2[x{{\sin }^{-1}}x+\sqrt{1-x^{2}}]+c $
D) $ 3[x{{\sin }^{-1}}x+\sqrt{1-x^{2}}]+c $
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Answer:
Correct Answer: D
Solution:
Put $ x=\sin \theta \Rightarrow dx=\cos \theta ,d\theta , $ therefore $ \int_{{}}^{{}}{{{\sin }^{-1}}(3x-4x^{3})},dx=\int_{{}}^{{}}{{{\sin }^{-1}}(\sin 3\theta )\cos \theta ,d\theta } $ $ =\int_{{}}^{{}}{3\theta \cos \theta ,d\theta }=3{ \theta \sin \theta -\int_{{}}^{{}}{\sin \theta ,d\theta } } $ $ =3{ \theta \sin \theta +\cos \theta }+c=3{ x{{\sin }^{-1}}x+\sqrt{1-x^{2}} }+c. $
BETA
