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Evaluate:$\int \limits_0^ {\pi/4} \sin 2x \sin 3x dx$

1 Answer

  • $(i)\; \int \limits_a^b f(x)dx =F(b)-F(a)$
  • $(ii)\; \cos (A-B)-\cos (A+B)= 2 \sin \bigg(\large \frac{A+B}{2}\bigg). \sin \bigg(\frac{A-B}{2}\bigg)$
  • $(iii)\; \int \cos x dx=\sin x$
Given $I= \int \limits_0^{\pi/4} \sin 2x \sin 3x dx$
Consider $\large\frac{1}{2}$$ [2 \sin 2x\sin 3x],$ this is of the form $ 2 \sin \bigg(\large\frac{A+B}{2}\bigg). $$\sin \bigg(\large\frac{A-B}{2}\bigg)=$$\cos (A-B)- \cos (A+B)$
Therefore $\large\frac{1}{2}$$ [2 \sin 2x\sin 3x]=\large\frac{1}{2}$$[(\cos x-\cos 5x )]$
Therefore $I=\large\frac{1}{2}$$\int \limits_0^{\pi/4} [\cos x-\cos 5x]dx$
$=\large\frac{1}{2}$$\bigg\{ \int \limits_0^{\pi/4} \cos x dx-\int \limits_0^{\pi/4} \cos x 5x\bigg\}$
On integrating we get
$I= \large\frac{1}{2} $$ \bigg[\sin x\bigg]_0^{\pi/4}$$-\bigg[\large\frac{\sin 5x}{5}\bigg]_0^{\pi/4}$
On applying limits,
$I=\large\frac{1}{2} $$\bigg\{ [\sin 4x-\sin 0]-\large\frac{1}{5} $$[\sin 5(\pi/4)-\sin 0]\bigg\}$
$\sin \pi/4=\sin 5 \pi/4 =\large\frac{1}{\sqrt 2}$ and $ \sin 0=0$
Therefore $ I=\large\frac{1}{2}$$ \bigg[\sin \pi/4-\frac{1}{5} \sin \bigg(\large\frac{5 \pi}{5}\bigg)\bigg]$
$=\large\frac{1}{2} \bigg[ \frac{1}{\sqrt 2}-\frac{1}{5} \times \frac{1}{\sqrt 2}\bigg]$
$=\large\frac{1}{2} \bigg(\large\frac{6}{5 \sqrt 2}\bigg)$
$=\large\frac{3}{5 \sqrt 2}$
Multiply and divide by $\sqrt 2$
$=\large\frac{3}{5 \sqrt 2} \times \frac{\sqrt 2}{\sqrt 2} =\frac{3 \sqrt 2}{10}$
answered Apr 25, 2013 by meena.p