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Question

Mathematics Question on integral

∫023dx4+9x2∫_0^\frac{2}{3} \frac{dx}{4+9x^2} equals

A

π6\frac{π}{6}

B

π12\frac{π}{12}

C

π24\frac{π}{24}

D

π4\frac{π}{4}

Answer

π24\frac{π}{24}

Explanation

Solution

∫∫dx4+9x2∫ \frac{dx}{4+9x^2}=∫dx(2)2+(3x)2∫\frac{dx}{(2)^2+(3x)^2}

put 3x=t⇒3dx=dt3x=t⇒3dx=dt

∴∫dx(2)2+(3x)2∫\frac{dx}{(2)^2+(3x)^2}=13∫dt(2)2+t2\frac{1}{3} ∫\frac{dt}{(2)^2+t^2}

=13[12tan−1t2]\frac{1}{3}[\frac{1}{2}tan^{-1}\frac{t}{2}]

=16\frac{1}{6}tan-1(3x2)(\frac{3x}{2})

=F(x)

By second fundamental theorem of calculus,we obtain

∫023dx4+9x2∫_0^\frac{2}{3} \frac{dx}{4+9x^2}=F(23)−F(0)F(\frac{2}{3})-F(0)

=16tan−1(32.23)−16tan−10=\frac{1}{6}tan^{-1}(\frac{3}{2}.\frac{2}{3})-\frac{1}{6}tan^{-1} 0

=16\frac{1}{6}tan-1 1-0

=16×π4\frac{1}{6}×\frac{π}{4}

=π24\frac{π}{24}

Hence,the correct Answer is C.