Question

Evaluate the definite integral
$\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x+\cos x}{9+16\sin 2x}}dx$

Answer 1

- Hint: First of all, assume sin x – cos x = t. Differentiate it and square it to get the value (sin x + cos x) dx and sin 2x in terms of t. Now, transfer the given integral in terms of t and use $\int{\dfrac{dx}{{{a}^{2}}-{{x}^{2}}}}=\log \left| \dfrac{a+x}{a-x} \right|+c$ to solve the given integral. Substitute the new limits to get the required answer.

Complete step-by-step solution -

In this question, we have to solve the given definite integral
$\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x+\cos x}{9+16\sin 2x}}dx$
Let us consider the integral given in the question.
$I=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x+\cos x}{9+16\sin 2x}}dx.....\left( i \right)$
Let us assume sin x – cos x = t….(ii)
We know that,
$\dfrac{d}{dx}\left( \sin x \right)=\cos x$
$\dfrac{d}{dx}\left( \cos x \right)=-\sin x$
So, by differentiating both the sides of equation (ii), we get,
$\left( \cos x+\sin x \right)=\dfrac{dt}{dx}$
$\left( \cos x+\sin x \right)dx=dt...\left( iii \right)$
Also, by squaring both the sides of equation (ii), we get,
${{\left( \sin x-\cos x \right)}^{2}}={{t}^{2}}$
We know that,
${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
By using this, we get,
${{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x={{t}^{2}}$
By using this ${{\sin }^{2}}x+{{\cos }^{2}}x=1\text{ and }2\sin x\cos x=\sin 2x$, we get,
$1-\sin 2x={{t}^{2}}$
$\sin 2x=1-{{t}^{2}}.....\left( iv \right)$
So, by substituting the value of (sin x + cos x)dx and sin 2x from equation (iii) and (iv) in equation (i), we get,
$I=\int\limits_{A}^{B}{\dfrac{dt}{9+16\left( 1-{{t}^{2}} \right)}}$
Let us find the value of the units A and B by substituting x = 0 and $x=\dfrac{\pi }{4}$ in equation (ii).
$A=\sin \left( 0 \right)-\cos \left( 0 \right)$
A = 0 – 1
A = – 1
$B=\sin \left( \dfrac{\pi }{4} \right)-\cos \left( \dfrac{\pi }{4} \right)$
$B=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}=0$
So, we get,
$I=\int\limits_{-1}^{0}{\dfrac{dt}{9+16\left( 1-{{t}^{2}} \right)}}$
$I=\int\limits_{-1}^{0}{\dfrac{dt}{9+16-16{{t}^{2}}}}$
$I=\int\limits_{-1}^{0}{\dfrac{dt}{25-16{{t}^{2}}}}$
By taking out 16 common from the denominator, we get,
$I=\dfrac{1}{16}\int\limits_{-1}^{0}{\dfrac{dt}{\dfrac{25}{16}-{{t}^{2}}}}$
$I=\dfrac{1}{16}\int\limits_{-1}^{0}{\dfrac{dt}{{{\left( \dfrac{5}{4} \right)}^{2}}-{{\left( t \right)}^{2}}}}$
We can see that the above integral is of the form $\int{\dfrac{dx}{{{a}^{2}}-{{x}^{2}}}}$ and we know that $\int{\dfrac{dx}{{{a}^{2}}-{{x}^{2}}}}=\dfrac{1}{2a}\log \left| \dfrac{a+x}{a-x} \right|+c$. By using this, we get,
$I=\dfrac{1}{16}\left[ \dfrac{1}{2.\left( \dfrac{5}{4} \right)}\log \left| \dfrac{\dfrac{5}{4}+t}{\dfrac{5}{4}-t} \right| \right]_{-1}^{0}$
$I=\dfrac{1}{4}\left[ \dfrac{1}{10}\log \left| \dfrac{5+4t}{5-4t} \right| \right]_{-1}^{0}$
$I=\dfrac{1}{40}\left[ \log \left| \dfrac{5+4t}{5-4t} \right| \right]_{-1}^{0}$
$I=\dfrac{1}{40}\left[ \log \left| \dfrac{5+4\left( 0 \right)}{5-4\left( 0 \right)} \right|-\log \left| \dfrac{5+4\left( -1 \right)}{5-4\left( -1 \right)} \right| \right]$
$I=\dfrac{1}{40}\left[ \log \left| \dfrac{5}{5} \right|-\log \left| \dfrac{1}{9} \right| \right]$
$I=\dfrac{1}{40}\left[ \log 1-\log \left( \dfrac{1}{9} \right) \right]$
We know that log 1 = 0. By using this, we get,
$I=\dfrac{1}{40}\left( -\log \dfrac{1}{9} \right)$
We know that $n\log m=\log {{m}^{n}}$. By using this, we get,
$I=\dfrac{1}{40}\left[ \log {{\left( \dfrac{1}{9} \right)}^{-1}} \right]$
$I=\dfrac{1}{40}\log 9$
Hence, we get the value of $\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x+\cos x}{9+16\sin 2x}}dx\text{ as }\dfrac{1}{40}\log 9$

Note: In this question, many students solve the integral correctly but forget to change the limits that is like in the above equation, they forget to transform the limits 0 and $\dfrac{\pi }{4}$ into 1 and 0 respectively and get the wrong answer even after doing the question correctly. So, this must be taken care of in the case of definite integrals. Also, students must remember the formula for general integrals to easily solve the question without making it very lengthy.