Question

If we have the integration as $\int{{{\sin }^{4}}x\times {{\cos }^{4}}x\times dx=\dfrac{1}{128}\left[ ax-\sin 4x+\dfrac{1}{8}\sin 8x \right]+C}$ then the value of ‘a’ is equal to:

Answer 1

Hint: Integrate the given integral using the trigonometric identities of $\sin 2x$ and $\cos 2x$ given as $\sin 2x=2\sin x\cos x$ and $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x$. Convert the degree of the given equation as linear and hence integrate by using $\int{\cos x=\sin x}$. Then equate with the given value and obtain the desired result.

Complete step-by-step solution -
Here, we have
$\int{{{\sin }^{4}}x\times {{\cos }^{4}}x\times dx=\dfrac{1}{128}\left[ ax-\sin 4x+\dfrac{1}{8}\sin 8x \right]+C}$
Hence, we need to solve the LHS of the above equation and compare the result with the RHS to get the value of ‘a’.
So, we can write
$LHS=\int{{{\sin }^{4}}x\cdot {{\cos }^{4}}xdx}$
$\Rightarrow LHS=\int{{{\left( \sin x\cos x \right)}^{4}}dx}$ ….........................................................(i)
Now, multiply and divide by ‘2’ in the bracket of equation (i), we get
$LHS={{\int{\left( \dfrac{2\sin x\cos x}{2} \right)}}^{4}}dx$
$LHS=\dfrac{1}{{{2}^{4}}}{{\int{\left( 2\sin x\cos x \right)}}^{4}}dx$
Now, we can replace $2\sin x\cos x$ by $\sin 2x$ using the identity $\sin 2x=2\sin x\cos x$ from the trigonometric function.
Hence, LHS can be re-written as
$LHS=\dfrac{1}{{{2}^{4}}}\int{{{\left( \sin 2x \right)}^{4}}dx}$
$\Rightarrow LHS=\dfrac{1}{{{2}^{4}}}\int{{{\sin }^{4}}2xdx}$ ……………………………………………….(ii)
Now, we can use the identity of$\cos 2x$ in terms of sine function can be given as,
$\cos 2x=1-2{{\sin }^{2}}x$
${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$
Here, we can replace x by 2x in the above relation and hence, we get
${{\sin }^{2}}2x=\dfrac{1-\cos 4x}{2}$ ……………………………………………(iii)
Now, we can put value of ${{\sin }^{2}}2x$ by using relation given in equation (iii), in the equation (ii), we get
$LHS=\dfrac{1}{{{2}^{4}}}\int{{{\left( {{\sin }^{2}}2x \right)}^{2}}dx}$
$LHS=\dfrac{1}{16}\int{{{\left( \dfrac{1-\cos 4x}{2} \right)}^{2}}dx}$
LHS$=\dfrac{1}{16\times 4}\int{{{\left( 1-\cos 4x \right)}^{2}}dx}$
Now, we can expand ${{\left( 1-\cos 2x \right)}^{2}}$ by using algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, we get
LHS$=\dfrac{1}{64}\int{\left( 1+{{\cos }^{2}}4x-2\cos 4x \right)dx}$
LHS$=\dfrac{1}{64}\left[ \int{1dx+\int{{{\cos }^{2}}4x}dx-2\int{\cos 4xdx}} \right]$ …………………………………………(iv)
Now, we can replace ${{\cos }^{2}}2x$ using the identity of $\cos 2x=2{{\cos }^{2}}x-1$ in following way : -
${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}$
Now, put $x=4x$ in the above equation, we get
${{\cos }^{2}}4x=\dfrac{1+\cos 8x}{2}$ ………………………………………………………..(v)
Hence, put value of ${{\cos }^{2}}2x$ from equation (v) in equation (iv), we get
LHS $=\dfrac{1}{64}\left[ \int{1dx+\int{\dfrac{1+\cos 8x}{2}dx-2\int{\cos 4xdx}}} \right]$
Now, we know the$\int{{{x}^{n}}}$ ,$\int{\cos x}$ can be given as $\int{{{x}^{n}}dx=\dfrac{{{x}^{n}}+1}{n+1}}$ and $\int{\cos xdx=\sin x}$
Hence, we can simplify LHS as
LHS$=\dfrac{1}{64}\left[ \int{{{x}^{0}}dx}+\dfrac{1}{2}\int{{{x}^{0}}dx}+\dfrac{1}{2}\int{\cos 8xdx-2\int{\cos 4xdx}} \right]$
LHS$=\dfrac{1}{64}\left[ x+\dfrac{1}{2}\left( x \right)+\dfrac{1}{2}\dfrac{\sin 8x}{8}-2\dfrac{\sin 4x}{4} \right]+c$
LHS$=\dfrac{1}{64}\left[ \dfrac{3x}{2}+\dfrac{1}{2}\dfrac{\sin 8x}{8}-\dfrac{\sin 4x}{2} \right]+c$
Now, take $\dfrac{1}{2}$ as common from bracket, we get
LHS$=\dfrac{1}{128}\left[ 3x+\dfrac{\sin 8x}{8}-\sin 4x \right]+c$
$\Rightarrow LHS=\dfrac{1}{128}\left[ 3x-\sin 4x+\dfrac{\sin 8x}{8} \right]+c$ ………………………………………….(vi)
Now consider the given equation,
$\int{{{\sin }^{4}}x{{\cos }^{4}}xdx=\dfrac{1}{128}\left[ ax-\sin 4x+\dfrac{1}{8}\sin 8x \right]+c}$
Now, compare the above equation with equation (vi), we get
$a=3$
Therefore, the answer is ‘3’.

Note: Another approach for finding value of ‘a’ would be that we can differentiate the given equation on both sides, and compare the coefficients. So, we can get after differentiating as : -
${{\sin }^{4}}x{{\cos }^{4}}x=\dfrac{1}{128}\left[ a-4\cos 4x+\dfrac{8}{8}\cos 8x \right]$
${{\sin }^{4}}x{{\cos }^{4}}x=\dfrac{1}{128}\left[ a-4\cos 4x+\cos 8x \right]$
$8{{\left( 2\sin x\cos x \right)}^{4}}=a-4\cos 4x+\cos 8x$
$8{{\left( {{\sin }^{2}}2x \right)}^{2}}=a-4\cos 4x+\cos 8x$
$8{{\left( \dfrac{1-\cos 4x}{2} \right)}^{2}}=a-4\cos 4x+\cos 8x$
$2{{\left( 1-\cos 4x \right)}^{2}}=a-4\cos 4x+\cos 8x$
$2\left( 1+{{\cos }^{2}}4x-2\cos 4x \right)=a-4\cos 4x+\cos 8x$
$2\left[ 1+{{\left( \dfrac{1-\cos 8x}{2} \right)}^{2}}-2\cos 4x \right]=a-4\cos 4x+8\cos 8x$
Now, simplify the above equation and get the value of ‘a’.
One may use another approach for finding $\int{{{\sin }^{4}}x{{\cos }^{4}}xdx}$ and may get values not in terms of $\sin 4x$ and $\sin 8x$. So, first we need to convert them in the form of the values given in RHS or we can simplify RHS as well. So, don’t confuse the terms. One may get value of $\int{{{\sin }^{4}}x{{\cos }^{4}}xdx}$ not in $\sin 4x$ and $\sin 8x$ .