Question

If $$I_n=\int {(\sin x)}^{n}dx,n\in N.$$ Then, $$5I_4-6I_6$$ is equal to:
$$\text{(a) }\sin x.{(\cos x)}^5+C$$
$$\left(\text{b}\right)\sin 2x.\cos 2x+C$$
$$\left(\text{c}\right){\displaystyle \frac{\sin 2x}{8}}[{\cos }^{2}x+1-2\cos 2x]+C$$
$$\left(\text{d}\right){\displaystyle \frac{\sin 2x}{8}}[{\cos }^{2}x+1+2\cos 2x]+C$$

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 $$I_4\text{ and }{I}_6\text{ in }{5I}_4-6I_6$$ 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,
$$I_n=\int {(\sin x)}^{n}dx$$
When we put n = 4 in the above equation, we get,
$$I_4=\int {(\sin x)}^{4}dx$$
Now, we have to find the value of $$I_4.$$ As we know that the direct integration of $${(\sin x)}^4$$ does not exist, so we will try to reduce the power. For this, we will write $${(\sin x)}^4\text{ as }{\left({\sin }^{2}x\right)}^2$$ 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,
$$\cos 2\theta =1-2{\sin }^2\theta$$
$$\Rightarrow 1-\cos 2\theta =2{\sin }^2\theta$$
$$\Rightarrow {\sin }^2\theta ={\displaystyle \frac{1-\cos 2\theta }{2}}$$
By using the above identity, we will get,
$$I_4={\int \left({\displaystyle \frac{1-\cos 2x}{2}}\right)}^{2}dx$$
$$\Rightarrow I_4={\int {\displaystyle \frac{(1-\cos 2x)}{4}}}^{2}dx$$
$$\Rightarrow I_4={\displaystyle \frac{1}{4}}\int {(1-\cos 2x)}^{2}dx$$
Now, we will use the identity,
$${(a-b)}^2=a^2+b^2-2ab.$$
Thus, we will get,
$$\Rightarrow I_4={\displaystyle \frac{1}{4}}\int (1+{\cos }^{2}2x-2\cos 2x)dx.....\left(i\right)$$
Now, here we will use another trigonometric identity:
$$\cos 2\theta =2{\cos }^2\theta -1$$
$$\Rightarrow 2{\cos }^2\theta =1+\cos 2\theta$$
$$\Rightarrow {\cos }^2\theta ={\displaystyle \frac{1+\cos 2\theta }{2}}$$
Putting $$\theta =2x,$$ we will get,
$$\Rightarrow {\cos }^{2}2x={\displaystyle \frac{1+\cos 4x}{2}}.....\left(ii\right)$$
Putting the value of $${\cos }^{2}2x$$ from (ii) to (i), we will get,
$$\Rightarrow I_4={\displaystyle \frac{1}{4}}\int (1+{\displaystyle \frac{1+\cos 4x}{2}}-2\cos 2x)dx$$
$$\Rightarrow I_4={\displaystyle \frac{1}{4}}\int (1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{\cos 4x}{2}}-2\cos 2x)dx$$
$$\Rightarrow I_4={\displaystyle \frac{1}{4}}\int ({\displaystyle \frac{3}{2}}+{\displaystyle \frac{\cos 4x}{2}}-2\cos 2x)dx$$
$$\Rightarrow I_4={\displaystyle \frac{1}{4}}\int {\displaystyle \frac{3}{2}}dx+{\displaystyle \frac{1}{4}}\int {\displaystyle \frac{\cos 4x}{2}}dx-{\displaystyle \frac{1}{4}}\int 2\cos 2xdx$$
Now, here we will use the following integration formulas,
$$\int adx=ax+C$$
$$\int \cos adx={\displaystyle \frac{\sin ax}{a}}+C$$
$$\Rightarrow I_4=[{\displaystyle \frac{1}{4}}\left({\displaystyle \frac{3}{2}}x\right)+C_1]+[{\displaystyle \frac{1}{4}}\left({\displaystyle \frac{\sin 4x}{4}}\right)\left({\displaystyle \frac{1}{2}}\right)+C_2]-[{\displaystyle \frac{1}{4}}\left({\displaystyle \frac{2\sin 2x}{2}}\right)+C_3]$$
$$\Rightarrow I_4={\displaystyle \frac{3}{8}}x+C_1+{\displaystyle \frac{\sin 4x}{32}}+C_2-{\displaystyle \frac{\sin 2x}{4}}+C_3$$
Considering the constants $$C_1=C_2=C_3=C_4,$$ we get,
$$\Rightarrow I_4={\displaystyle \frac{3x}{8}}+{\displaystyle \frac{\sin 4x}{32}}-{\displaystyle \frac{\sin 2x}{4}}+C_4.....\left(iii\right)$$
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,
$$\sin 3\theta =3\sin \theta -4{\sin }^3\theta$$
$$\Rightarrow 4{\sin }^3\theta =3\sin \theta -\sin 3\theta$$
$$\Rightarrow {\sin }^3\theta ={\displaystyle \frac{3\sin \theta -\sin 3\theta }{4}}$$
Thus, after using this identity, we will get,
$$\Rightarrow I_6={\int \left({\displaystyle \frac{3\sin x-\sin 3x}{4}}\right)}^{2}dx$$
$$\Rightarrow I_6={\displaystyle \frac{1}{16}}{\int (3\sin x-\sin 3x)}^{2}dx$$
Now, we will use the identity,
$${(a-b)}^2=a^2+b^2-2ab$$
Thus, we will get,
$$\Rightarrow I_6={\displaystyle \frac{1}{16}}\int (9{\sin }^{2}x+{\sin }^{2}3x-6\sin x\sin 3x)dx$$
Now, here we will use two trigonometric identities,
$$\cos 2\theta =1-2{\sin }^2\theta$$
$$\Rightarrow {\sin }^2\theta ={\displaystyle \frac{1-\cos 2\theta }{2}}$$
Similarly,
$${\sin }^{2}3\theta ={\displaystyle \frac{1-\cos 6\theta }{2}}$$
Also, 2 sin A sin B = cos (A – B) – cos (A + B)
Thus, we will get,
$$\Rightarrow I_6={\displaystyle \frac{1}{16}}\int [{\displaystyle \frac{9}{2}}(1-\cos 2x)+{\displaystyle \frac{1}{2}}(1-\cos 6x)-3(2\sin x\sin 3x)]dx$$
$$\Rightarrow I_6={\displaystyle \frac{1}{16}}\int [{\displaystyle \frac{9}{2}}-{\displaystyle \frac{9\cos 2x}{2}}+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{\cos 6x}{2}}-3(\cos (x-3x)-\cos (x+3x))]dx$$
$$\Rightarrow I_6={\displaystyle \frac{1}{16}}\int [5-{\displaystyle \frac{9\cos 2x}{2}}-{\displaystyle \frac{\cos 6x}{2}}-3(\cos 2x-\cos 4x)]dx$$
$$\Rightarrow I_6={\displaystyle \frac{1}{16}}\int [5-{\displaystyle \frac{15\cos 2x}{2}}+3\cos 4x-{\displaystyle \frac{\cos 6x}{2}}]dx$$
$$\Rightarrow I_6={\displaystyle \frac{1}{16}}\int 5dx-{\displaystyle \frac{1}{16}}\int {\displaystyle \frac{15\cos 2x}{2}}dx+{\displaystyle \frac{1}{16}}\int 3\cos 4xdx-{\displaystyle \frac{1}{16}}\int {\displaystyle \frac{\cos 6x}{2}}dx$$
Now, we will use some integration formulas here,
$$\int adx=ax+c$$
$$\int \cos axdx={\displaystyle \frac{\sin ax}{a}}+c$$
Thus, we will get,
$$\Rightarrow I_6=[{\displaystyle \frac{1}{16}}\left(5x\right)+C_5]-[{\displaystyle \frac{1}{16}}\left({\displaystyle \frac{15\sin 2x}{4}}\right)+C_6]+[{\displaystyle \frac{1}{16}}\left({\displaystyle \frac{3\sin 4x}{4}}\right)+C_7]-[{\displaystyle \frac{1}{16}}\left({\displaystyle \frac{\cos 6x}{12}}\right)+C_8]$$
$$\Rightarrow I_6={\displaystyle \frac{5x}{16}}+C_5-{\displaystyle \frac{15\sin 2x}{64}}+C_6+{\displaystyle \frac{3\sin 4x}{64}}+C_7-{\displaystyle \frac{\sin 6x}{192}}+C_8$$
Considering the constants $$C_5=C_6=C_7=C_8=C_9,$$ we get,
$$\Rightarrow I_6={\displaystyle \frac{5x}{16}}-{\displaystyle \frac{15\sin 2x}{64}}+{\displaystyle \frac{3\sin 4x}{64}}-{\displaystyle \frac{\sin 6x}{192}}+C_9.....\left(iv\right)$$
Now, we have to find the value of $$5I_4-6I_6.$$ Thus, we will get,
$$5I_4-6I_6=5[{\displaystyle \frac{3x}{8}}+{\displaystyle \frac{\sin 4x}{32}}-{\displaystyle \frac{\sin 2x}{4}}+C_4]-6[{\displaystyle \frac{5x}{16}}-{\displaystyle \frac{15\sin 2x}{64}}+{\displaystyle \frac{3\sin 4x}{64}}-{\displaystyle \frac{\sin 6x}{192}}+C_9]$$
$$5I_4-6I_6={\displaystyle \frac{15x}{8}}+{\displaystyle \frac{5\sin 4x}{32}}-{\displaystyle \frac{5\sin 2x}{4}}+5C_4-{\displaystyle \frac{15x}{8}}+{\displaystyle \frac{45\sin 2x}{32}}-{\displaystyle \frac{9\sin 4x}{32}}+{\displaystyle \frac{\sin 6x}{32}}-6C_9$$
Taking the constant, $$5C_4-6C_9=C,$$ we get,
$$5I_4-6I_6={\displaystyle \frac{-1}{8}}\sin 4x+{\displaystyle \frac{5}{32}}\sin 2x+{\displaystyle \frac{\sin 6x}{32}}+C.....\left(v\right)$$
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,
$$5I_4-6I_6={\displaystyle \frac{-1}{8}}[2\sin 2x\cos 2x]+{\displaystyle \frac{5}{32}}\sin 2x+{\displaystyle \frac{1}{32}}[3\sin 2x-4{\sin }^{3}2x]+C$$
$$5I_4-6I_6={\displaystyle \frac{-\sin 2x\cos 2x}{4}}+{\displaystyle \frac{5\sin 2x}{32}}+{\displaystyle \frac{3\sin 2x}{32}}-{\displaystyle \frac{4{\sin }^{3}2x}{32}}+C$$
$$5I_4-6I_6={\displaystyle \frac{-\sin 2x\cos 2x}{4}}+{\displaystyle \frac{\sin 2x}{4}}-{\displaystyle \frac{{\sin }^{2}2x}{8}}+C$$
$$5I_4-6I_6={\displaystyle \frac{\sin 2x}{8}}[-2\cos 2x+2-{\sin }^{2}2x]+C$$
$$5I_4-6I_6={\displaystyle \frac{\sin 2x}{8}}[-2\cos 2x+2-(1-{\cos }^{2}2x)]+C$$
$$5I_4-6I_6={\displaystyle \frac{\sin 2x}{8}}[1+{\cos }^{2}2x-2\cos 2x]+C$$
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,
$$\int {\sin }^{n}xdx={\displaystyle \frac{-1}{n}}{\sin }^{(n-1)}x\cos x+{\displaystyle \frac{(n-1)}{n}}\int {\sin }^{(n-2)}xdx$$
$$\Rightarrow n\int {\sin }^{n}xdx=-{\sin }^{(n-1)}x\cos x+(n-1)\int {\sin }^{(n-2)}xdx$$
$$\Rightarrow n{I}_n=-{\sin }^{(n-1)}x\cos x+(n-1)I_{n-2}$$
Now, we will put n = 6. Thus, we will get,
$$\Rightarrow 6I_6=-{\sin }^{(6-1)}x\cos x+(6-1)I_{6-2}$$
$$\Rightarrow 6I_6=-{\sin }^{5}x\cos x+5I_4$$
$$\Rightarrow 5I_4-6I_6={\sin }^{5}x\cos x$$
$$\Rightarrow 5I_4-6I_6={\sin }^{4}x(\sin x\cos x)$$
Now, we will use sin 2x = 2 sin x cos x. Thus, we will get,
$$\Rightarrow 5I_4-6I_6={\sin }^{4}x\left({\displaystyle \frac{\sin 2x}{2}}\right)={\left({\sin }^{2}x\right)}^2\left({\displaystyle \frac{\sin 2x}{2}}\right)$$
Now, we will use the identity,
$${\sin }^{2}x={\displaystyle \frac{1-\cos 2x}{2}}$$
Thus, we get,
$$\Rightarrow 5I_4-6I_6={\displaystyle \frac{{(1-\cos 2x)}^2}{4}}\left({\displaystyle \frac{\sin 2x}{2}}\right)$$
$$\Rightarrow 5I_4-6I_6={\displaystyle \frac{\sin 2x}{8}}[1+{\cos }^{2}2x-2\cos 2x]$$