Question

Find the value of the integration \[\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}} {\left\{ {\dfrac{{1 + \sin 2x + \cos 2x}}{{\sin x + \cos x}}} \right\}dx} \] .
A. 16
B. 8
C. 4
D. 1

Answer 1

Hints First substitute \[1 + \cos 2x\] by the formula \[2{\cos ^2}x\] and \[\sin 2x\] by the formula \[2\sin x\cos x\] in the numerator of the given fraction. Then factor out the term \[(\cos x + \sin x)\] from the obtained expression. Then cancel out the term \[(\cos x + \sin x)\] from the numerator and the denominator of the obtained expression. Then integrate the simplified expression to obtain the required answer.

Formula used
\[1 + \cos 2A = 2{\cos ^2}A\]
\[\sin 2A = 2\sin A\cos A\]
\[\int {\cos x} = \sin x\]

Complete step by step solution
The given expression is,
\[\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}} {\left\{ {\dfrac{{1 + \sin 2x + \cos 2x}}{{\sin x + \cos x}}} \right\}dx} \]
\[ = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}} {\left\{ {\dfrac{{\left( {1 + \cos 2x} \right) + \sin 2x}}{{\sin x + \cos x}}} \right\}dx} \]
Apply the formula \[1 + \cos 2A = 2{\cos ^2}A\] and \[\sin 2A = 2\sin A\cos A\] in the given expression for further calculation.
\[ = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}} {\left\{ {\dfrac{{2{{\cos }^2}x + 2\sin x\cos x}}{{\sin x + \cos x}}} \right\}dx} \]
\[ = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}} {\left\{ {\dfrac{{2\cos x(\cos x + \sin x)}}{{\sin x + \cos x}}} \right\}dx} \]
\[ = 2\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}} {\cos xdx} \]
\[ = 2\left[ {\sin x} \right]_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}\]
\[ = 2\left[ {\sin \dfrac{\pi }{2} - \sin \dfrac{\pi }{6}} \right]\]
\[ = 2\left[ {1 - \dfrac{1}{2}} \right]\]
\[ = 2 \times \dfrac{1}{2}\]
\[ = 1\]
The correct option is D.

Note Sometime students did not apply the formula to the given expression as shown and directly apply \[\sin 2A = 2\sin A\cos A\] and \[{\cos ^2}A - {\sin ^2}A = \cos 2A\]. Then in the final result they get the answer as 1 but this process is lengthy.