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Question: Find the value of \(\int {\dfrac{{\cos x - \sin x}}{{\sqrt {8 - \sin 2x} }}} dx\)...

Find the value of ∫cos⁡x−sin⁡x8−sin⁡2xdx\int {\dfrac{{\cos x - \sin x}}{{\sqrt {8 - \sin 2x} }}} dx

Explanation

Solution

According to given in the question we have to determine the value of the given integration∫cos⁡x−sin⁡x8−sin⁡2xdx\int {\dfrac{{\cos x - \sin x}}{{\sqrt {8 - \sin 2x} }}} dx. So, first of all we have to let (cos⁡x+sin⁡x)(\cos x + \sin x) equal to t and then differentiate both side with respect to t and further, we used the formula of differentiation of sin⁡x\sin x and cos⁡x\cos x as mentioned below

Formula used:
⇒ddxsin⁡x=cos⁡x\Rightarrow \dfrac{d}{{dx}}\sin x = \cos x
⇒ddxcos⁡x=−sin⁡x\Rightarrow \dfrac{d}{{dx}}\cos x = - \sin x
Now, we have to use the formula of (a+b)2{(a + b)^2}as mentioned below:

⇒(a+b)2=a2+b2+2ab \Rightarrow {(a + b)^2} = {a^2} + {b^2} + 2ab

Complete step by step answer:
Step 1: First of all we have to let the given integration function equal to I as mentioned below:
⇒I=∫cos⁡x−sin⁡x8−sin⁡2xdx\Rightarrow I = \int {\dfrac{{\cos x - \sin x}}{{\sqrt {8 - \sin 2x} }}} dx…………………………………..(1)
Now, we have to let (cos⁡x+sin⁡x)(\cos x + \sin x) equal to t. As mentioned below:
⇒t=(cos⁡x+sin⁡x)\Rightarrow t = (\cos x + \sin x)……………………………………..(2)
Step 2: Now, We have to differentiate the above expression (2) with respect to t from both sides. As mentioned below:
⇒ddtt=ddt(cos⁡x−sin⁡x) ⇒1=ddtcos⁡x−ddtsin⁡x ⇒1=(−sin⁡x+cos⁡x)dxdt ⇒dt=dx(cos⁡x−sin⁡x).........................(3)  \Rightarrow \dfrac{d}{{dt}}t = \dfrac{d}{{dt}}(\cos x - \sin x) \\\ \Rightarrow 1 = \dfrac{d}{{dt}}\cos x - \dfrac{d}{{dt}}\sin x \\\ \Rightarrow 1 = ( - \sin x + \cos x)\dfrac{{dx}}{{dt}} \\\ \Rightarrow dt = dx(\cos x - \sin x).........................(3) \\\
Step 3: Now, We have to take the square of expression (2) from both sides. As mentioned below:
⇒t2=(cos⁡x+sin⁡x)2   \Rightarrow {t^2} = {(\cos x + \sin x)^2} \\\ \\\
Now, we have to use the formula of (a+b)2{(a + b)^2} as mentioned below:
Formula used:
⇒(a+b)2=a2+b2+2ab\Rightarrow {(a + b)^2} = {a^2} + {b^2} + 2ab
Step 4: Now, we get the expression as mentioned below:
⇒t2=cos⁡2x+sin⁡2x+2sin⁡x.cos⁡x   \Rightarrow {t^2} = {\cos ^2}x + {\sin ^2}x + 2\sin x.\cos x \\\ \\\
Now, we have to use the formula of cos⁡2x+sin⁡2x{\cos ^2}x + {\sin ^2}x as mentioned below:
Formula used:
cos⁡2x+sin⁡2x{\cos ^2}x + {\sin ^2}x= 1
⇒t2=1+2sin⁡x.cos⁡x   \Rightarrow {t^2} = 1 + 2\sin x.\cos x \\\ \\\
Now, we have to use the formula of 2sin⁡acos⁡a2\sin a\cos a as mentioned below:
Formula used:
2sin⁡acos⁡a=sin⁡2a2\sin a\cos a = \sin 2a
Now, we get the expression as mentioned below:
⇒t2=1+sin⁡2x ⇒t2−1=sin⁡2x.......................(4)   \Rightarrow {t^2} = 1 + \sin 2x \\\ \Rightarrow {t^2} - 1 = \sin 2x.......................(4) \\\ \\\
Now, we have to put the values of (3) and (4) in expression (1)
Step 5: Now, we get the expression as mentioned below:
⇒I=∫dt8−(t2−1) ⇒I=∫dt8−t2+1 ⇒I=∫dt9−t2 ⇒I=∫dt32−t2....................(5)  \Rightarrow I = \int {\dfrac{{dt}}{{\sqrt {8 - ({t^2} - 1)} }}} \\\ \Rightarrow I = \int {\dfrac{{dt}}{{\sqrt {8 - {t^2} + 1} }}} \\\ \Rightarrow I = \int {\dfrac{{dt}}{{\sqrt {9 - {t^2}} }}} \\\ \Rightarrow I = \int {\dfrac{{dt}}{{\sqrt {{3^2} - {t^2}} }}} ....................(5) \\\
Now, we use the integration formula of ∫1a2−sin⁡2xdx\int {\dfrac{1}{{\sqrt {{a^2} - {{\sin }^2}x} }}} dx that are mentioned below:
∫1a2−sin⁡2xdx\int {\dfrac{1}{{\sqrt {{a^2} - {{\sin }^2}x} }}} dx= sin⁡−1(xa){\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right)
Step 6: Now, the expression (5) will be after using the formula mentioned above:
⇒I=sin⁡−1(t3)+C\Rightarrow I = {\sin ^{ - 1}}\left( {\dfrac{t}{3}} \right) + C
Now, put the value of t in the above expression. We get:
⇒I=sin⁡−1(1+sin⁡2x3)+C\Rightarrow I = {\sin ^{ - 1}}\left( {\dfrac{{1 + \sin 2x}}{3}} \right) + C

The integrated values of the function ∫cos⁡x−sin⁡x8−sin⁡2xdx\int {\dfrac{{\cos x - \sin x}}{{\sqrt {8 - \sin 2x} }}} dx is sin⁡−1(1+sin⁡2x3)+C{\sin ^{ - 1}}\left( {\dfrac{{1 + \sin 2x}}{3}} \right) + C

Note: First of all we have to let the value of (sin⁡x+cos⁡x)(\sin x + \cos x) equals t so we can replace the value of sin⁡2x\sin 2x in terms of t after squaring the function which we let t.
Now, we integrated the function in terms of t and then using the formula of integration of ∫1a2−sin⁡2xdx\int {\dfrac{1}{{\sqrt {{a^2} - {{\sin }^2}x} }}} dx this function.