Question

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

Answer 1

Hint: To solve the question given above, we will first find out the values of \[{{I}_{4}}\text{ and }{{I}_{6}}\] by putting n = 4 and n = 6 respectively in the equation given. For the calculation of \[{{I}_{4}}\text{ and }{{I}_{6}},\] 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 \[{{I}_{4}}\text{ and }{{I}_{6}}\text{ in }{{ 5I }_{4}}-6{{I}_{6}}\] and then solve.

Complete step-by-step answer:
In the first step, we are going to calculate the value of \[{{I}_{4}}.\] We know that,
\[{{I}_{n}}=\int{{{\left( \sin x \right)}^{n}}dx}\]
When we put n = 4 in the above equation, we get,
\[{{I}_{4}}=\int{{{\left( \sin x \right)}^{4}}dx}\]
Now, we have to find the value of \[{{I}_{4}}.\] As we know that the direct integration of \[{{\left( \sin x \right)}^{4}}\] does not exist, so we will try to reduce the power. For this, we will write \[{{\left( \sin x \right)}^{4}}\text{ as }{{\left( {{\sin }^{2}}x \right)}^{2}}\] with the help of the identity \[{{\left( {{a}^{b}} \right)}^{c}}={{a}^{b\times c}}.\] Thus, we will get,
\[{{I}_{4}}=\int{{{\left( {{\sin }^{2}}x \right)}^{2}}dx}\]
Now, here we will use a trigonometric identity,
\[\cos 2\theta =1-2{{\sin }^{2}}\theta \]
\[\Rightarrow 1-\cos 2\theta =2{{\sin }^{2}}\theta \]
\[\Rightarrow {{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}\]
By using the above identity, we will get,
\[{{I}_{4}}={{\int{\left( \dfrac{1-\cos 2x}{2} \right)}}^{2}}dx\]
\[\Rightarrow {{I}_{4}}={{\int{\dfrac{\left( 1-\cos 2x \right)}{4}}}^{2}}dx\]
\[\Rightarrow {{I}_{4}}=\dfrac{1}{4}\int{{{\left( 1-\cos 2x \right)}^{2}}}dx\]
Now, we will use the identity,
\[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab.\]
Thus, we will get,
\[\Rightarrow {{I}_{4}}=\dfrac{1}{4}\int{\left( 1+{{\cos }^{2}}2x-2\cos 2x \right)}dx.....\left( i \right)\]
Now, here we will use another trigonometric identity:
\[\cos 2\theta =2{{\cos }^{2}}\theta -1\]
\[\Rightarrow 2{{\cos }^{2}}\theta =1+\cos 2\theta \]
\[\Rightarrow {{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2}\]
Putting \[\theta =2x,\] we will get,
\[\Rightarrow {{\cos }^{2}}2x=\dfrac{1+\cos 4x}{2}.....\left( ii \right)\]
Putting the value of \[{{\cos }^{2}}2x\] from (ii) to (i), we will get,
\[\Rightarrow {{I}_{4}}=\dfrac{1}{4}\int{\left( 1+\dfrac{1+\cos 4x}{2}-2\cos 2x \right)dx}\]
\[\Rightarrow {{I}_{4}}=\dfrac{1}{4}\int{\left( 1+\dfrac{1}{2}+\dfrac{\cos 4x}{2}-2\cos 2x \right)dx}\]
\[\Rightarrow {{I}_{4}}=\dfrac{1}{4}\int{\left( \dfrac{3}{2}+\dfrac{\cos 4x}{2}-2\cos 2x \right)dx}\]
\[\Rightarrow {{I}_{4}}=\dfrac{1}{4}\int{\dfrac{3}{2}dx}+\dfrac{1}{4}\int{\dfrac{\cos 4x}{2}dx}-\dfrac{1}{4}\int{2\cos 2xdx}\]
Now, here we will use the following integration formulas,
\[\int{adx}=ax+C\]
\[\int{\cos adx}=\dfrac{\sin ax}{a}+C\]
\[\Rightarrow {{I}_{4}}=\left[ \dfrac{1}{4}\left( \dfrac{3}{2}x \right)+{{C}_{1}} \right]+\left[ \dfrac{1}{4}\left( \dfrac{\sin 4x}{4} \right)\left( \dfrac{1}{2} \right)+{{C}_{2}} \right]-\left[ \dfrac{1}{4}\left( \dfrac{2\sin 2x}{2} \right)+{{C}_{3}} \right]\]
\[\Rightarrow {{I}_{4}}=\dfrac{3}{8}x+{{C}_{1}}+\dfrac{\sin 4x}{32}+{{C}_{2}}-\dfrac{\sin 2x}{4}+{{C}_{3}}\]
Considering the constants \[{{C}_{1}}={{C}_{2}}={{C}_{3}}={{C}_{4}},\] we get,
\[\Rightarrow {{I}_{4}}=\dfrac{3x}{8}+\dfrac{\sin 4x}{32}-\dfrac{\sin 2x}{4}+{{C}_{4}}.....\left( iii \right)\]
Now, similarly, we will calculate the value of \[{{I}_{6}}.\] Thus, we have,
\[{{I}_{6}}=\int{\left( {{\sin }^{6}}x \right)dx}\]
\[\Rightarrow {{I}_{6}}=\int{{{\left( {{\sin }^{3}}x \right)}^{2}}dx}\]
Now, we will use a trigonometric identity here,
\[\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta \]
\[\Rightarrow 4{{\sin }^{3}}\theta =3\sin \theta -\sin 3\theta \]
\[\Rightarrow {{\sin }^{3}}\theta =\dfrac{3\sin \theta -\sin 3\theta }{4}\]
Thus, after using this identity, we will get,
\[\Rightarrow {{I}_{6}}={{\int{\left( \dfrac{3\sin x-\sin 3x}{4} \right)}}^{2}}dx\]
\[\Rightarrow {{I}_{6}}=\dfrac{1}{16}{{\int{\left( 3\sin x-\sin 3x \right)}}^{2}}dx\]
Now, we will use the identity,
\[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]
Thus, we will get,
\[\Rightarrow {{I}_{6}}=\dfrac{1}{16}\int{\left( 9{{\sin }^{2}}x+{{\sin }^{2}}3x-6\sin x\sin 3x \right)}dx\]
Now, here we will use two trigonometric identities,
\[\cos 2\theta =1-2{{\sin }^{2}}\theta \]
\[\Rightarrow {{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}\]
Similarly,
\[{{\sin }^{2}}3\theta =\dfrac{1-\cos 6\theta }{2}\]
Also, 2 sin A sin B = cos (A – B) – cos (A + B)
Thus, we will get,
\[\Rightarrow {{I}_{6}}=\dfrac{1}{16}\int{\left[ \dfrac{9}{2}\left( 1-\cos 2x \right)+\dfrac{1}{2}\left( 1-\cos 6x \right)-3\left( 2\sin x\sin 3x \right) \right]}dx\]
\[\Rightarrow {{I}_{6}}=\dfrac{1}{16}\int{\left[ \dfrac{9}{2}-\dfrac{9\cos 2x}{2}+\dfrac{1}{2}-\dfrac{\cos 6x}{2}-3\left( \cos \left( x-3x \right)-\cos \left( x+3x \right) \right) \right]}\text{ }dx\]
\[\Rightarrow {{I}_{6}}=\dfrac{1}{16}\int{\left[ 5-\dfrac{9\cos 2x}{2}-\dfrac{\cos 6x}{2}-3\left( \cos 2x-\cos 4x \right) \right]}\text{ }dx\]
\[\Rightarrow {{I}_{6}}=\dfrac{1}{16}\int{\left[ 5-\dfrac{15\cos 2x}{2}+3\cos 4x-\dfrac{\cos 6x}{2} \right]}\text{ }dx\]
\[\Rightarrow {{I}_{6}}=\dfrac{1}{16}\int{5dx}-\dfrac{1}{16}\int{\dfrac{15\cos 2x}{2}}dx+\dfrac{1}{16}\int{3\cos 4xdx}-\dfrac{1}{16}\int{\dfrac{\cos 6x}{2}}dx\]
Now, we will use some integration formulas here,
\[\int{adx=ax+c}\]
\[\int{\cos axdx=\dfrac{\sin ax}{a}+c}\]
Thus, we will get,
\[\Rightarrow {{I}_{6}}=\left[ \dfrac{1}{16}\left( 5x \right)+{{C}_{5}} \right]-\left[ \dfrac{1}{16}\left( \dfrac{15\sin 2x}{4} \right)+{{C}_{6}} \right]+\left[ \dfrac{1}{16}\left( \dfrac{3\sin 4x}{4} \right)+{{C}_{7}} \right]-\left[ \dfrac{1}{16}\left( \dfrac{\cos 6x}{12} \right)+{{C}_{8}} \right]\]
\[\Rightarrow {{I}_{6}}=\dfrac{5x}{16}+{{C}_{5}}-\dfrac{15\sin 2x}{64}+{{C}_{6}}+\dfrac{3\sin 4x}{64}+{{C}_{7}}-\dfrac{\sin 6x}{192}+{{C}_{8}}\]
Considering the constants \[{{C}_{5}}={{C}_{6}}={{C}_{7}}={{C}_{8}}={{C}_{9}},\] we get,
\[\Rightarrow {{I}_{6}}=\dfrac{5x}{16}-\dfrac{15\sin 2x}{64}+\dfrac{3\sin 4x}{64}-\dfrac{\sin 6x}{192}+{{C}_{9}}.....\left( iv \right)\]
Now, we have to find the value of \[5{{I}_{4}}-6{{I}_{6}}.\] Thus, we will get,
\[5{{I}_{4}}-6{{I}_{6}}=5\left[ \dfrac{3x}{8}+\dfrac{\sin 4x}{32}-\dfrac{\sin 2x}{4}+{{C}_{4}} \right]-6\left[ \dfrac{5x}{16}-\dfrac{15\sin 2x}{64}+\dfrac{3\sin 4x}{64}-\dfrac{\sin 6x}{192}+{{C}_{9}} \right]\]
\[5{{I}_{4}}-6{{I}_{6}}=\dfrac{15x}{8}+\dfrac{5\sin 4x}{32}-\dfrac{5\sin 2x}{4}+5{{C}_{4}}-\dfrac{15x}{8}+\dfrac{45\sin 2x}{32}-\dfrac{9\sin 4x}{32}+\dfrac{\sin 6x}{32}-6{{C}_{9}}\]
Taking the constant, \[5{{C}_{4}}-6{{C}_{9}}=C,\] we get,
\[5{{I}_{4}}-6{{I}_{6}}=\dfrac{-1}{8}\sin 4x+\dfrac{5}{32}\sin 2x+\dfrac{\sin 6x}{32}+C.....\left( v \right)\]
Now, we will use two trigonometric identities here,
\[\sin 2x=2\sin x\cos x\]
\[\Rightarrow \sin 4x=2\sin 2x\cos 2x....\left( vi \right)\]
\[\sin 3x=3\sin x-4{{\sin }^{3}}x\]
\[\Rightarrow \sin 6x=3\sin 2x-4{{\sin }^{3}}\left( 2x \right).....\left( vii \right)\]
Putting the values of sin 4x and sin 6x from (vi) and (vii) to (v), we will get,
\[5{{I}_{4}}-6{{I}_{6}}=\dfrac{-1}{8}\left[ 2\sin 2x\cos 2x \right]+\dfrac{5}{32}\sin 2x+\dfrac{1}{32}\left[ 3\sin 2x-4{{\sin }^{3}}2x \right]+C\]
\[5{{I}_{4}}-6{{I}_{6}}=\dfrac{-\sin 2x\cos 2x}{4}+\dfrac{5\sin 2x}{32}+\dfrac{3\sin 2x}{32}-\dfrac{4{{\sin }^{3}}2x}{32}+C\]
\[5{{I}_{4}}-6{{I}_{6}}=\dfrac{-\sin 2x\cos 2x}{4}+\dfrac{\sin 2x}{4}-\dfrac{{{\sin }^{2}}2x}{8}+C\]
\[5{{I}_{4}}-6{{I}_{6}}=\dfrac{\sin 2x}{8}\left[ -2\cos 2x+2-{{\sin }^{2}}2x \right]+C\]
\[5{{I}_{4}}-6{{I}_{6}}=\dfrac{\sin 2x}{8}\left[ -2\cos 2x+2-\left( 1-{{\cos }^{2}}2x \right) \right]+C\]
\[5{{I}_{4}}-6{{I}_{6}}=\dfrac{\sin 2x}{8}\left[ 1+{{\cos }^{2}}2x-2\cos 2x \right]+C\]
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,
\[\int{{{\sin }^{n}}xdx}=\dfrac{-1}{n}{{\sin }^{\left( n-1 \right)}}x\cos x+\dfrac{\left( n-1 \right)}{n}\int{{{\sin }^{\left( n-2 \right)}}}xdx\]
\[\Rightarrow n\int{{{\sin }^{n}}xdx}=-{{\sin }^{\left( n-1 \right)}}x\cos x+\left( n-1 \right)\int{{{\sin }^{\left( n-2 \right)}}}xdx\]
\[\Rightarrow n{{I}_{n}}=-{{\sin }^{\left( n-1 \right)}}x\cos x+\left( n-1 \right){{I}_{n-2}}\]
Now, we will put n = 6. Thus, we will get,
\[\Rightarrow 6{{I}_{6}}=-{{\sin }^{\left( 6-1 \right)}}x\cos x+\left( 6-1 \right){{I}_{6-2}}\]
\[\Rightarrow 6{{I}_{6}}=-{{\sin }^{5}}x\cos x+5{{I}_{4}}\]
\[\Rightarrow 5{{I}_{4}}-6{{I}_{6}}={{\sin }^{5}}x\cos x\]
\[\Rightarrow 5{{I}_{4}}-6{{I}_{6}}={{\sin }^{4}}x\left( \sin x\cos x \right)\]
Now, we will use sin 2x = 2 sin x cos x. Thus, we will get,
\[\Rightarrow 5{{I}_{4}}-6{{I}_{6}}={{\sin }^{4}}x\left( \dfrac{\sin 2x}{2} \right)={{\left( {{\sin }^{2}}x \right)}^{2}}\left( \dfrac{\sin 2x}{2} \right)\]
Now, we will use the identity,
\[{{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}\]
Thus, we get,
\[\Rightarrow 5{{I}_{4}}-6{{I}_{6}}=\dfrac{{{\left( 1-\cos 2x \right)}^{2}}}{4}\left( \dfrac{\sin 2x}{2} \right)\]
\[\Rightarrow 5{{I}_{4}}-6{{I}_{6}}=\dfrac{\sin 2x}{8}\left[ 1+{{\cos }^{2}}2x-2\cos 2x \right]\]