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Question: The value of \(2\int {\sin x\cos ec4xdx} \) is equal to A) \(\dfrac{1}{{2\sqrt 2 }}\ln \left| {\d...

The value of 2∫sin⁡xcos⁡ec4xdx2\int {\sin x\cos ec4xdx} is equal to
A) 122ln⁡∣1+2sin⁡x1−2sin⁡x∣−14ln⁡∣1+sin⁡x1−sin⁡x∣+c\dfrac{1}{{2\sqrt 2 }}\ln \left| {\dfrac{{1 + \sqrt 2 \sin x}}{{1 - \sqrt 2 \sin x}}} \right| - \dfrac{1}{4}\ln \left| {\dfrac{{1 + \sin x}}{{1 - \sin x}}} \right| + c
B) 122ln⁡∣1+2sin⁡x1−2sin⁡x∣−12ln⁡∣1+sin⁡xcos⁡x∣+c\dfrac{1}{{2\sqrt 2 }}\ln \left| {\dfrac{{1 + \sqrt 2 \sin x}}{{1 - \sqrt 2 \sin x}}} \right| - \dfrac{1}{2}\ln \left| {\dfrac{{1 + \sin x}}{{\cos x}}} \right| + c
C) 122ln⁡∣1−2sin⁡x1+2sin⁡x∣−14ln⁡∣1+sin⁡x1−sin⁡x∣+c\dfrac{1}{{2\sqrt 2 }}\ln \left| {\dfrac{{1 - \sqrt 2 \sin x}}{{1 + \sqrt 2 \sin x}}} \right| - \dfrac{1}{4}\ln \left| {\dfrac{{1 + \sin x}}{{1 - \sin x}}} \right| + c
D) 122ln⁡∣1−2sin⁡x1+2sin⁡x∣+12ln⁡∣1+sin⁡xcos⁡x∣+c\dfrac{1}{{2\sqrt 2 }}\ln \left| {\dfrac{{1 - \sqrt 2 \sin x}}{{1 + \sqrt 2 \sin x}}} \right| + \dfrac{1}{2}\ln \left| {\dfrac{{1 + \sin x}}{{\cos x}}} \right| + c

Explanation

Solution

We will first write cosec x in terms of sin x and after that we will expand the term using trigonometric identity: sin 2x = 2 sin x cos x. We will use another trigonometric identity:
cos⁡2x=1−2sin⁡2x\cos 2x=1-2\sin ^{2}x afterwards. Using the identity cos⁡2x=1−sin⁡2x\cos ^{2}x=1-\sin ^{2}x, we will simplify it further. Now, we will put sin x = t and hence, cos x dx = dt. We will solve it further for the terms to reduce in some known form of integral function. We will use the formula of integrals as:
∫dx(a2−x2)=12aln⁡∣a+xa−x∣\int {\dfrac{{dx}}{{\left( {{a^2} - {x^2}} \right)}} = \dfrac{1}{{2a}}\ln \left| {\dfrac{{a + x}}{{a - x}}} \right|} +c and then we will find the required value of the given integral.

Complete step by step solution:
We are given an integral: 2∫sin⁡xcos⁡ec4xdx2\int {\sin x\cos ec4xdx} .
Let us first convert cosec x in to sin x using the formula: cosec x = 1sin⁡x\dfrac{1}{{\sin x}}.
We can write the given integral as:
⇒2∫sin⁡xsin⁡4xdx\Rightarrow 2\int {\dfrac{{\sin x}}{{\sin 4x}}dx}
Now, expanding sin 4x using the trigonometric identity: sin 2x = 2 sin x cos x, here x = 2x.
⇒2∫sin⁡x2sin⁡2xcos⁡2xdx\Rightarrow 2\int {\dfrac{{\sin x}}{{2\sin 2x\cos 2x}}dx}
Again, simplifying the denominator using the trigonometric identity: sin 2x = 2 sin x cos x, here x = x.
⇒2∫sin⁡x4sin⁡xcos⁡xcos⁡2xdx\Rightarrow 2\int {\dfrac{{\sin x}}{{4\sin x\cos x\cos 2x}}dx}
Using the trigonometric identity: cos⁡2x=1−2sin⁡2x\cos 2x=1-2\sin ^{2}x, we get
⇒12∫1cos⁡xcos⁡2xdx ⇒12∫1cos⁡x(1−2sin⁡2x)dx  \Rightarrow \dfrac{1}{2}\int {\dfrac{1}{{\cos x\cos 2x}}dx} \\\ \Rightarrow \dfrac{1}{2}\int {\dfrac{1}{{\cos x\left( {1 - 2{{\sin }^2}x} \right)}}dx} \\\
Multiplying and dividing by cos x, we get
⇒12∫cos⁡xcos⁡2x(1−2sin⁡2x)dx\Rightarrow \dfrac{1}{2}\int {\dfrac{{\cos x}}{{{{\cos }^2}x\left( {1 - 2{{\sin }^2}x} \right)}}dx}
Using the trigonometric identity: cos⁡2x=1−sin⁡2x\cos ^{2}x=1-\sin ^{2}x, we get
⇒12∫cos⁡x(1−sin⁡2x)(1−2sin⁡2x)dx\Rightarrow \dfrac{1}{2}\int {\dfrac{{\cos x}}{{\left( {1 - {{\sin }^2}x} \right)\left( {1 - 2{{\sin }^2}x} \right)}}dx}
Now, put the value of sin x = t and hence on differentiating both sides, we get cos x dx = dt. Substituting these values in the integral obtained above, we get
⇒12∫1(1−t2)(1−2t2)dt\Rightarrow \dfrac{1}{2}\int {\dfrac{1}{{\left( {1 - {t^2}} \right)\left( {1 - 2{t^2}} \right)}}} dt
We can write the numerator 1=2(1−t2)−(1−2t2)1=2\left( 1-t^2 \right)-\left(1-2t^2 \right). Putting this value in the numerator, we get
⇒12∫2(1−t2)−(1−2t2)(1−t2)(1−2t2)dt\Rightarrow \dfrac{1}{2}\int {\dfrac{{2\left( {1 - {t^2}} \right) - \left( {1 - 2{t^2}} \right)}}{{\left( {1 - {t^2}} \right)\left( {1 - 2{t^2}} \right)}}} dt
We can separate the integral as:
⇒12∫2(1−t2)(1−t2)(1−2t2)dt−12∫(1−2t2)(1−t2)(1−2t2)dt\Rightarrow \dfrac{1}{2}\int {\dfrac{{2\left( {1 - {t^2}} \right)}}{{\left( {1 - {t^2}} \right)\left( {1 - 2{t^2}} \right)}}} dt - \dfrac{1}{2}\int {\dfrac{{\left( {1 - 2{t^2}} \right)}}{{\left( {1 - {t^2}} \right)\left( {1 - 2{t^2}} \right)}}} dt
⇒12∫2(1−2t2)dt−12∫1(1−t2)dt\Rightarrow \dfrac{1}{2}\int {\dfrac{2}{{\left( {1 - 2{t^2}} \right)}}} dt - \dfrac{1}{2}\int {\dfrac{1}{{\left( {1 - {t^2}} \right)}}} dt
Now, using the integral formula: ∫dx(a2−x2)=12aln⁡∣a+xa−x∣\int {\dfrac{{dx}}{{\left( {{a^2} - {x^2}} \right)}} = \dfrac{1}{{2a}}\ln \left| {\dfrac{{a + x}}{{a - x}}} \right|} + c in the above integrals, we can write
⇒12(222ln⁡∣1+2t1−2t∣)−12(12ln⁡∣1+t1−t∣)\Rightarrow \dfrac{1}{2}\left( {\dfrac{2}{{2\sqrt 2 }}\ln \left| {\dfrac{{1 + \sqrt 2 t}}{{1 - \sqrt 2 t}}} \right|} \right) - \dfrac{1}{2}\left( {\dfrac{1}{2}\ln \left| {\dfrac{{1 + t}}{{1 - t}}} \right|} \right)+ c
Substituting the value of t = sin x in the equation, we get
⇒122ln⁡∣1+2sin⁡x1−2sin⁡x∣−14ln⁡∣1+sin⁡x1−sin⁡x∣+c\Rightarrow \dfrac{1}{{2\sqrt 2 }}\ln \left| {\dfrac{{1 + \sqrt 2 \sin x}}{{1 - \sqrt 2 \sin x}}} \right| - \dfrac{1}{4}\ln \left| {\dfrac{{1 + \sin x}}{{1 - \sin x}}} \right| + c

Hence, we can say that option(A) is correct.

Note:
For any integration problem our first approach should be to resolve the problem into simpler one by any means. In this question we have used trigonometric identity to resolve the integral into the known formula then we can directly apply the formula as we did in our solution.