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Question: If \[{{I}_{n}}=\int{{{\left( \sin x \right)}^{n}}dx,n\in N.}\] Then, \[5{{I}_{4}}-6{{I}_{6}}\] is eq...

If In=∫(sin⁡x)ndx,n∈N.{{I}_{n}}=\int{{{\left( \sin x \right)}^{n}}dx,n\in N.} Then, 5I4−6I65{{I}_{4}}-6{{I}_{6}} is equal to:
(a) sin⁡x.(cos⁡x)5+C\text{(a) }\sin x.{{\left( \cos x \right)}^{5}}+C
(b) sin⁡2x.cos⁡2x+C\left( \text{b} \right)\text{ }\sin 2x.\cos 2x+C
(c) sin⁡2x8[cos⁡2x+1−2cos⁡2x]+C\left( \text{c} \right)\text{ }\dfrac{\sin 2x}{8}\left[ {{\cos }^{2}}x+1-2\cos 2x \right]+C
(d) sin⁡2x8[cos⁡2x+1+2cos⁡2x]+C\left( \text{d} \right)\text{ }\dfrac{\sin 2x}{8}\left[ {{\cos }^{2}}x+1+2\cos 2x \right]+C

Explanation

Solution

Hint: To solve the question given above, we will first find out the values of I4 and I6{{I}_{4}}\text{ and }{{I}_{6}} by putting n = 4 and n = 6 respectively in the equation given. For the calculation of I4 and I6,{{I}_{4}}\text{ and }{{I}_{6}}, our main aim will be to convert higher powers of the sine function to the sine functions or cosine functions with single power. Then we will put the values of I4 and I6 in 5I4−6I6{{I}_{4}}\text{ and }{{I}_{6}}\text{ in }{{ 5I }_{4}}-6{{I}_{6}} and then solve.

Complete step-by-step answer:
In the first step, we are going to calculate the value of I4.{{I}_{4}}. We know that,
In=∫(sin⁡x)ndx{{I}_{n}}=\int{{{\left( \sin x \right)}^{n}}dx}
When we put n = 4 in the above equation, we get,
I4=∫(sin⁡x)4dx{{I}_{4}}=\int{{{\left( \sin x \right)}^{4}}dx}
Now, we have to find the value of I4.{{I}_{4}}. As we know that the direct integration of (sin⁡x)4{{\left( \sin x \right)}^{4}} does not exist, so we will try to reduce the power. For this, we will write (sin⁡x)4 as (sin⁡2x)2{{\left( \sin x \right)}^{4}}\text{ as }{{\left( {{\sin }^{2}}x \right)}^{2}} with the help of the identity (ab)c=ab×c.{{\left( {{a}^{b}} \right)}^{c}}={{a}^{b\times c}}. Thus, we will get,
I4=∫(sin⁡2x)2dx{{I}_{4}}=\int{{{\left( {{\sin }^{2}}x \right)}^{2}}dx}
Now, here we will use a trigonometric identity,
cos⁡2θ=1−2sin⁡2θ\cos 2\theta =1-2{{\sin }^{2}}\theta
⇒1−cos⁡2θ=2sin⁡2θ\Rightarrow 1-\cos 2\theta =2{{\sin }^{2}}\theta
⇒sin⁡2θ=1−cos⁡2θ2\Rightarrow {{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}
By using the above identity, we will get,
I4=∫(1−cos⁡2x2)2dx{{I}_{4}}={{\int{\left( \dfrac{1-\cos 2x}{2} \right)}}^{2}}dx
⇒I4=∫(1−cos⁡2x)42dx\Rightarrow {{I}_{4}}={{\int{\dfrac{\left( 1-\cos 2x \right)}{4}}}^{2}}dx
⇒I4=14∫(1−cos⁡2x)2dx\Rightarrow {{I}_{4}}=\dfrac{1}{4}\int{{{\left( 1-\cos 2x \right)}^{2}}}dx
Now, we will use the identity,
(a−b)2=a2+b2−2ab.{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab.
Thus, we will get,
⇒I4=14∫(1+cos⁡22x−2cos⁡2x)dx.....(i)\Rightarrow {{I}_{4}}=\dfrac{1}{4}\int{\left( 1+{{\cos }^{2}}2x-2\cos 2x \right)}dx.....\left( i \right)
Now, here we will use another trigonometric identity:
cos⁡2θ=2cos⁡2θ−1\cos 2\theta =2{{\cos }^{2}}\theta -1
⇒2cos⁡2θ=1+cos⁡2θ\Rightarrow 2{{\cos }^{2}}\theta =1+\cos 2\theta
⇒cos⁡2θ=1+cos⁡2θ2\Rightarrow {{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2}
Putting θ=2x,\theta =2x, we will get,
⇒cos⁡22x=1+cos⁡4x2.....(ii)\Rightarrow {{\cos }^{2}}2x=\dfrac{1+\cos 4x}{2}.....\left( ii \right)
Putting the value of cos⁡22x{{\cos }^{2}}2x from (ii) to (i), we will get,
⇒I4=14∫(1+1+cos⁡4x2−2cos⁡2x)dx\Rightarrow {{I}_{4}}=\dfrac{1}{4}\int{\left( 1+\dfrac{1+\cos 4x}{2}-2\cos 2x \right)dx}
⇒I4=14∫(1+12+cos⁡4x2−2cos⁡2x)dx\Rightarrow {{I}_{4}}=\dfrac{1}{4}\int{\left( 1+\dfrac{1}{2}+\dfrac{\cos 4x}{2}-2\cos 2x \right)dx}
⇒I4=14∫(32+cos⁡4x2−2cos⁡2x)dx\Rightarrow {{I}_{4}}=\dfrac{1}{4}\int{\left( \dfrac{3}{2}+\dfrac{\cos 4x}{2}-2\cos 2x \right)dx}
⇒I4=14∫32dx+14∫cos⁡4x2dx−14∫2cos⁡2xdx\Rightarrow {{I}_{4}}=\dfrac{1}{4}\int{\dfrac{3}{2}dx}+\dfrac{1}{4}\int{\dfrac{\cos 4x}{2}dx}-\dfrac{1}{4}\int{2\cos 2xdx}
Now, here we will use the following integration formulas,
∫adx=ax+C\int{adx}=ax+C
∫cos⁡adx=sin⁡axa+C\int{\cos adx}=\dfrac{\sin ax}{a}+C
⇒I4=[14(32x)+C1]+[14(sin⁡4x4)(12)+C2]−[14(2sin⁡2x2)+C3]\Rightarrow {{I}_{4}}=\left[ \dfrac{1}{4}\left( \dfrac{3}{2}x \right)+{{C}_{1}} \right]+\left[ \dfrac{1}{4}\left( \dfrac{\sin 4x}{4} \right)\left( \dfrac{1}{2} \right)+{{C}_{2}} \right]-\left[ \dfrac{1}{4}\left( \dfrac{2\sin 2x}{2} \right)+{{C}_{3}} \right]
⇒I4=38x+C1+sin⁡4x32+C2−sin⁡2x4+C3\Rightarrow {{I}_{4}}=\dfrac{3}{8}x+{{C}_{1}}+\dfrac{\sin 4x}{32}+{{C}_{2}}-\dfrac{\sin 2x}{4}+{{C}_{3}}
Considering the constants C1=C2=C3=C4,{{C}_{1}}={{C}_{2}}={{C}_{3}}={{C}_{4}}, we get,
⇒I4=3x8+sin⁡4x32−sin⁡2x4+C4.....(iii)\Rightarrow {{I}_{4}}=\dfrac{3x}{8}+\dfrac{\sin 4x}{32}-\dfrac{\sin 2x}{4}+{{C}_{4}}.....\left( iii \right)
Now, similarly, we will calculate the value of I6.{{I}_{6}}. Thus, we have,
I6=∫(sin⁡6x)dx{{I}_{6}}=\int{\left( {{\sin }^{6}}x \right)dx}
⇒I6=∫(sin⁡3x)2dx\Rightarrow {{I}_{6}}=\int{{{\left( {{\sin }^{3}}x \right)}^{2}}dx}
Now, we will use a trigonometric identity here,
sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta
⇒4sin⁡3θ=3sin⁡θ−sin⁡3θ\Rightarrow 4{{\sin }^{3}}\theta =3\sin \theta -\sin 3\theta
⇒sin⁡3θ=3sin⁡θ−sin⁡3θ4\Rightarrow {{\sin }^{3}}\theta =\dfrac{3\sin \theta -\sin 3\theta }{4}
Thus, after using this identity, we will get,
⇒I6=∫(3sin⁡x−sin⁡3x4)2dx\Rightarrow {{I}_{6}}={{\int{\left( \dfrac{3\sin x-\sin 3x}{4} \right)}}^{2}}dx
⇒I6=116∫(3sin⁡x−sin⁡3x)2dx\Rightarrow {{I}_{6}}=\dfrac{1}{16}{{\int{\left( 3\sin x-\sin 3x \right)}}^{2}}dx
Now, we will use the identity,
(a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab
Thus, we will get,
⇒I6=116∫(9sin⁡2x+sin⁡23x−6sin⁡xsin⁡3x)dx\Rightarrow {{I}_{6}}=\dfrac{1}{16}\int{\left( 9{{\sin }^{2}}x+{{\sin }^{2}}3x-6\sin x\sin 3x \right)}dx
Now, here we will use two trigonometric identities,
cos⁡2θ=1−2sin⁡2θ\cos 2\theta =1-2{{\sin }^{2}}\theta
⇒sin⁡2θ=1−cos⁡2θ2\Rightarrow {{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}
Similarly,
sin⁡23θ=1−cos⁡6θ2{{\sin }^{2}}3\theta =\dfrac{1-\cos 6\theta }{2}
Also, 2 sin A sin B = cos (A – B) – cos (A + B)
Thus, we will get,
⇒I6=116∫[92(1−cos⁡2x)+12(1−cos⁡6x)−3(2sin⁡xsin⁡3x)]dx\Rightarrow {{I}_{6}}=\dfrac{1}{16}\int{\left[ \dfrac{9}{2}\left( 1-\cos 2x \right)+\dfrac{1}{2}\left( 1-\cos 6x \right)-3\left( 2\sin x\sin 3x \right) \right]}dx
⇒I6=116∫[92−9cos⁡2x2+12−cos⁡6x2−3(cos⁡(x−3x)−cos⁡(x+3x))] dx\Rightarrow {{I}_{6}}=\dfrac{1}{16}\int{\left[ \dfrac{9}{2}-\dfrac{9\cos 2x}{2}+\dfrac{1}{2}-\dfrac{\cos 6x}{2}-3\left( \cos \left( x-3x \right)-\cos \left( x+3x \right) \right) \right]}\text{ }dx
⇒I6=116∫[5−9cos⁡2x2−cos⁡6x2−3(cos⁡2x−cos⁡4x)] dx\Rightarrow {{I}_{6}}=\dfrac{1}{16}\int{\left[ 5-\dfrac{9\cos 2x}{2}-\dfrac{\cos 6x}{2}-3\left( \cos 2x-\cos 4x \right) \right]}\text{ }dx
⇒I6=116∫[5−15cos⁡2x2+3cos⁡4x−cos⁡6x2] dx\Rightarrow {{I}_{6}}=\dfrac{1}{16}\int{\left[ 5-\dfrac{15\cos 2x}{2}+3\cos 4x-\dfrac{\cos 6x}{2} \right]}\text{ }dx
⇒I6=116∫5dx−116∫15cos⁡2x2dx+116∫3cos⁡4xdx−116∫cos⁡6x2dx\Rightarrow {{I}_{6}}=\dfrac{1}{16}\int{5dx}-\dfrac{1}{16}\int{\dfrac{15\cos 2x}{2}}dx+\dfrac{1}{16}\int{3\cos 4xdx}-\dfrac{1}{16}\int{\dfrac{\cos 6x}{2}}dx
Now, we will use some integration formulas here,
∫adx=ax+c\int{adx=ax+c}
∫cos⁡axdx=sin⁡axa+c\int{\cos axdx=\dfrac{\sin ax}{a}+c}
Thus, we will get,
⇒I6=[116(5x)+C5]−[116(15sin⁡2x4)+C6]+[116(3sin⁡4x4)+C7]−[116(cos⁡6x12)+C8]\Rightarrow {{I}_{6}}=\left[ \dfrac{1}{16}\left( 5x \right)+{{C}_{5}} \right]-\left[ \dfrac{1}{16}\left( \dfrac{15\sin 2x}{4} \right)+{{C}_{6}} \right]+\left[ \dfrac{1}{16}\left( \dfrac{3\sin 4x}{4} \right)+{{C}_{7}} \right]-\left[ \dfrac{1}{16}\left( \dfrac{\cos 6x}{12} \right)+{{C}_{8}} \right]
⇒I6=5x16+C5−15sin⁡2x64+C6+3sin⁡4x64+C7−sin⁡6x192+C8\Rightarrow {{I}_{6}}=\dfrac{5x}{16}+{{C}_{5}}-\dfrac{15\sin 2x}{64}+{{C}_{6}}+\dfrac{3\sin 4x}{64}+{{C}_{7}}-\dfrac{\sin 6x}{192}+{{C}_{8}}
Considering the constants C5=C6=C7=C8=C9,{{C}_{5}}={{C}_{6}}={{C}_{7}}={{C}_{8}}={{C}_{9}}, we get,
⇒I6=5x16−15sin⁡2x64+3sin⁡4x64−sin⁡6x192+C9.....(iv)\Rightarrow {{I}_{6}}=\dfrac{5x}{16}-\dfrac{15\sin 2x}{64}+\dfrac{3\sin 4x}{64}-\dfrac{\sin 6x}{192}+{{C}_{9}}.....\left( iv \right)
Now, we have to find the value of 5I4−6I6.5{{I}_{4}}-6{{I}_{6}}. Thus, we will get,
5I4−6I6=5[3x8+sin⁡4x32−sin⁡2x4+C4]−6[5x16−15sin⁡2x64+3sin⁡4x64−sin⁡6x192+C9]5{{I}_{4}}-6{{I}_{6}}=5\left[ \dfrac{3x}{8}+\dfrac{\sin 4x}{32}-\dfrac{\sin 2x}{4}+{{C}_{4}} \right]-6\left[ \dfrac{5x}{16}-\dfrac{15\sin 2x}{64}+\dfrac{3\sin 4x}{64}-\dfrac{\sin 6x}{192}+{{C}_{9}} \right]
5I4−6I6=15x8+5sin⁡4x32−5sin⁡2x4+5C4−15x8+45sin⁡2x32−9sin⁡4x32+sin⁡6x32−6C95{{I}_{4}}-6{{I}_{6}}=\dfrac{15x}{8}+\dfrac{5\sin 4x}{32}-\dfrac{5\sin 2x}{4}+5{{C}_{4}}-\dfrac{15x}{8}+\dfrac{45\sin 2x}{32}-\dfrac{9\sin 4x}{32}+\dfrac{\sin 6x}{32}-6{{C}_{9}}
Taking the constant, 5C4−6C9=C,5{{C}_{4}}-6{{C}_{9}}=C, we get,
5I4−6I6=−18sin⁡4x+532sin⁡2x+sin⁡6x32+C.....(v)5{{I}_{4}}-6{{I}_{6}}=\dfrac{-1}{8}\sin 4x+\dfrac{5}{32}\sin 2x+\dfrac{\sin 6x}{32}+C.....\left( v \right)
Now, we will use two trigonometric identities here,
sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x
⇒sin⁡4x=2sin⁡2xcos⁡2x....(vi)\Rightarrow \sin 4x=2\sin 2x\cos 2x....\left( vi \right)
sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x=3\sin x-4{{\sin }^{3}}x
⇒sin⁡6x=3sin⁡2x−4sin⁡3(2x).....(vii)\Rightarrow \sin 6x=3\sin 2x-4{{\sin }^{3}}\left( 2x \right).....\left( vii \right)
Putting the values of sin 4x and sin 6x from (vi) and (vii) to (v), we will get,
5I4−6I6=−18[2sin⁡2xcos⁡2x]+532sin⁡2x+132[3sin⁡2x−4sin⁡32x]+C5{{I}_{4}}-6{{I}_{6}}=\dfrac{-1}{8}\left[ 2\sin 2x\cos 2x \right]+\dfrac{5}{32}\sin 2x+\dfrac{1}{32}\left[ 3\sin 2x-4{{\sin }^{3}}2x \right]+C
5I4−6I6=−sin⁡2xcos⁡2x4+5sin⁡2x32+3sin⁡2x32−4sin⁡32x32+C5{{I}_{4}}-6{{I}_{6}}=\dfrac{-\sin 2x\cos 2x}{4}+\dfrac{5\sin 2x}{32}+\dfrac{3\sin 2x}{32}-\dfrac{4{{\sin }^{3}}2x}{32}+C
5I4−6I6=−sin⁡2xcos⁡2x4+sin⁡2x4−sin⁡22x8+C5{{I}_{4}}-6{{I}_{6}}=\dfrac{-\sin 2x\cos 2x}{4}+\dfrac{\sin 2x}{4}-\dfrac{{{\sin }^{2}}2x}{8}+C
5I4−6I6=sin⁡2x8[−2cos⁡2x+2−sin⁡22x]+C5{{I}_{4}}-6{{I}_{6}}=\dfrac{\sin 2x}{8}\left[ -2\cos 2x+2-{{\sin }^{2}}2x \right]+C
5I4−6I6=sin⁡2x8[−2cos⁡2x+2−(1−cos⁡22x)]+C5{{I}_{4}}-6{{I}_{6}}=\dfrac{\sin 2x}{8}\left[ -2\cos 2x+2-\left( 1-{{\cos }^{2}}2x \right) \right]+C
5I4−6I6=sin⁡2x8[1+cos⁡22x−2cos⁡2x]+C5{{I}_{4}}-6{{I}_{6}}=\dfrac{\sin 2x}{8}\left[ 1+{{\cos }^{2}}2x-2\cos 2x \right]+C
Hence, the option (c) is the right answer.

Note: We can also use the reduction formula to solve this question. Reduction formula for integral of the sine function is,
∫sin⁡nxdx=−1nsin⁡(n−1)xcos⁡x+(n−1)n∫sin⁡(n−2)xdx\int{{{\sin }^{n}}xdx}=\dfrac{-1}{n}{{\sin }^{\left( n-1 \right)}}x\cos x+\dfrac{\left( n-1 \right)}{n}\int{{{\sin }^{\left( n-2 \right)}}}xdx
⇒n∫sin⁡nxdx=−sin⁡(n−1)xcos⁡x+(n−1)∫sin⁡(n−2)xdx\Rightarrow n\int{{{\sin }^{n}}xdx}=-{{\sin }^{\left( n-1 \right)}}x\cos x+\left( n-1 \right)\int{{{\sin }^{\left( n-2 \right)}}}xdx
⇒nIn=−sin⁡(n−1)xcos⁡x+(n−1)In−2\Rightarrow n{{I}_{n}}=-{{\sin }^{\left( n-1 \right)}}x\cos x+\left( n-1 \right){{I}_{n-2}}
Now, we will put n = 6. Thus, we will get,
⇒6I6=−sin⁡(6−1)xcos⁡x+(6−1)I6−2\Rightarrow 6{{I}_{6}}=-{{\sin }^{\left( 6-1 \right)}}x\cos x+\left( 6-1 \right){{I}_{6-2}}
⇒6I6=−sin⁡5xcos⁡x+5I4\Rightarrow 6{{I}_{6}}=-{{\sin }^{5}}x\cos x+5{{I}_{4}}
⇒5I4−6I6=sin⁡5xcos⁡x\Rightarrow 5{{I}_{4}}-6{{I}_{6}}={{\sin }^{5}}x\cos x
⇒5I4−6I6=sin⁡4x(sin⁡xcos⁡x)\Rightarrow 5{{I}_{4}}-6{{I}_{6}}={{\sin }^{4}}x\left( \sin x\cos x \right)
Now, we will use sin 2x = 2 sin x cos x. Thus, we will get,
⇒5I4−6I6=sin⁡4x(sin⁡2x2)=(sin⁡2x)2(sin⁡2x2)\Rightarrow 5{{I}_{4}}-6{{I}_{6}}={{\sin }^{4}}x\left( \dfrac{\sin 2x}{2} \right)={{\left( {{\sin }^{2}}x \right)}^{2}}\left( \dfrac{\sin 2x}{2} \right)
Now, we will use the identity,
sin⁡2x=1−cos⁡2x2{{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}
Thus, we get,
⇒5I4−6I6=(1−cos⁡2x)24(sin⁡2x2)\Rightarrow 5{{I}_{4}}-6{{I}_{6}}=\dfrac{{{\left( 1-\cos 2x \right)}^{2}}}{4}\left( \dfrac{\sin 2x}{2} \right)
⇒5I4−6I6=sin⁡2x8[1+cos⁡22x−2cos⁡2x]\Rightarrow 5{{I}_{4}}-6{{I}_{6}}=\dfrac{\sin 2x}{8}\left[ 1+{{\cos }^{2}}2x-2\cos 2x \right]