Answer
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Hint: Divide the numerator and denominator of the expression by \[{{\cos }^{4}}x\] and convert the expression in terms of ‘tan’ and ‘sec’ and then put ‘tan x = t’ and use the formula \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C\] and get the answer in terms of ‘t’. Then replace ‘t’ by ‘tan x’ and simplify the equation to get the final answer.
Complete step by step answer:
To solve the given expression we will write it down first and assume it as ‘L’
\[\therefore L=\int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}}dx}\] …………………………………………………… (1)
As we see the above equation is in the form of ‘sin’ and ‘cos’ and to solve this we have to convert it in ‘tan’ so that we can solve it. If we add the powers of the term \[{{\sin }^{3}}x\times {{\cos }^{5}}x\] we will get, 3 + 5 = 8 and if we divide the term by \[{{\cos }^{8}}x\] we will get the term completely converted to ‘tan’.
To divide the term \[{{\sin }^{3}}x\times {{\cos }^{5}}x\] by \[{{\cos }^{8}}x\] we have to divide \[\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}\] by \[{{\cos }^{4}}x\] as \[\sqrt{{{\cos }^{8}}x}={{\cos }^{4}}x\].
Therefore dividing the above equation by \[{{\cos }^{4}}x\] to both the numerator and denominator we will get,
\[\therefore L=\int{\dfrac{\dfrac{1}{{{\cos }^{4}}x}}{\dfrac{\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}}{{{\cos }^{4}}x}}dx}\]
As we know that \[\dfrac{1}{{{\cos }^{4}}x}\] is equal to \[{{\sec }^{4}}x\] therefore above equation will become,
\[\therefore L=\int{\dfrac{{{\sec }^{4}}x}{\dfrac{\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}}{{{\cos }^{4}}x}}dx}\]
Also we know that \[{{\cos }^{4}}x=\sqrt{{{\cos }^{8}}x}\] therefore by replacing \[{{\cos }^{4}}x\] by \[\sqrt{{{\cos }^{8}}x}\] in the above equation we will get,
\[\therefore L=\int{\dfrac{{{\sec }^{4}}x}{\dfrac{\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}}{\sqrt{{{\cos }^{8}}x}}}dx}\]
Above equation can also be written as,
\[\therefore L=\int{\dfrac{{{\sec }^{4}}x}{\sqrt{\dfrac{{{\sin }^{3}}x\times {{\cos }^{5}}x}{{{\cos }^{8}}x}}}dx}\]
\[\therefore L=\int{\dfrac{{{\sec }^{4}}x}{\sqrt{\dfrac{{{\sin }^{3}}x\times {{\cos }^{5}}x}{{{\cos }^{\left( 3+5 \right)}}x}}}dx}\]
Now to proceed further in the solution we should know the formula given below,
Formula:
\[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\]
Referring to above formula we can replace \[{{\cos }^{\left( 3+5 \right)}}x\] by \[{{\cos }^{3}}x{{\cos }^{5}}x\] in ‘L’ therefore we will get,
\[\therefore L=\int{\dfrac{{{\sec }^{4}}x}{\sqrt{\dfrac{{{\sin }^{3}}x\times {{\cos }^{5}}x}{{{\cos }^{3}}x{{\cos }^{5}}x}}}dx}\]
As we all know that \[\dfrac{\sin x}{\cos x}=\tan x\] therefore above equation will become,
\[\therefore L=\int{\dfrac{{{\sec }^{4}}x}{\sqrt{{{\tan }^{3}}x}}dx}\]
We can also write the above equation as,
\[\therefore L=\int{\dfrac{{{\sec }^{2}}x\times {{\sec }^{2}}x}{\sqrt{{{\tan }^{3}}x}}dx}\]
As we all know the identity i.e. \[1+{{\tan }^{2}}x={{\sec }^{2}}x\] therefore above equation will become,
\[\therefore L=\int{\dfrac{\left( 1+{{\tan }^{2}}x \right)\times {{\sec }^{2}}x}{\sqrt{{{\tan }^{3}}x}}dx}\]
Now we will put, tan x = t in the above equation, ………………………………………………… (2)
Therefore, \[{{\sec }^{2}}xdx=dt\] ……………………………………………………………………………………. (3)
Also, as tan x = t
Squaring on both sides we will get,
\[{{\tan }^{2}}x={{t}^{2}}\] ………………………………………………….. (4)
If we put the values of equation (2), equation (3) and equation (4) in ‘L’ we will get,
\[\therefore L=\int{\dfrac{1+{{\operatorname{t}}^{2}}}{\sqrt{{{\operatorname{t}}^{3}}}}dt}\]
Above equation can also be written as,
\[\therefore L=\int{\dfrac{1+{{\operatorname{t}}^{2}}}{{{\operatorname{t}}^{\dfrac{3}{2}}}}dt}\]
If we separate the denominator in the above equation we will get,
\[\therefore L=\int{\left[ \dfrac{1}{{{\operatorname{t}}^{\dfrac{3}{2}}}}+\dfrac{{{\operatorname{t}}^{2}}}{{{\operatorname{t}}^{\dfrac{3}{2}}}} \right]dt}\]
\[\therefore L=\int{\left[ {{\operatorname{t}}^{-\dfrac{3}{2}}}+{{\operatorname{t}}^{2-\dfrac{3}{2}}} \right]dt}\]
\[\therefore L=\int{\left[ {{\operatorname{t}}^{-\dfrac{3}{2}}}+{{\operatorname{t}}^{\dfrac{4-3}{2}}} \right]dt}\]
\[\therefore L=\int{\left[ {{\operatorname{t}}^{-\dfrac{3}{2}}}+{{\operatorname{t}}^{\dfrac{1}{2}}} \right]dt}\]
Now to proceed further in the solution we should know the formula given below,
Formula:
\[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C\]
By using above formula in ‘L’ we will get,
\[\therefore L=\dfrac{{{\operatorname{t}}^{-\dfrac{3}{2}+1}}}{-\dfrac{3}{2}+1}+\dfrac{{{\operatorname{t}}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}+C\]
\[\therefore L=\dfrac{{{\operatorname{t}}^{1-\dfrac{3}{2}}}}{1-\dfrac{3}{2}}+\dfrac{{{\operatorname{t}}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}+C\]
\[\therefore L=\dfrac{{{\operatorname{t}}^{\dfrac{2-3}{2}}}}{\dfrac{2-3}{2}}+\dfrac{{{\operatorname{t}}^{\dfrac{1+2}{2}}}}{\dfrac{1+2}{2}}+C\]
\[\therefore L=\dfrac{{{\operatorname{t}}^{\dfrac{-1}{2}}}}{\dfrac{-1}{2}}+\dfrac{{{\operatorname{t}}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+C\]
\[\therefore L={{\operatorname{t}}^{\dfrac{-1}{2}}}\times \dfrac{-2}{1}+{{\operatorname{t}}^{\dfrac{3}{2}}}\times \dfrac{2}{3}+C\]
If we substitute the value of ‘t’ from equation (2) in the above equation we will get,
\[\therefore L=-2{{\tan }^{\dfrac{-1}{2}}}x+\dfrac{2}{3}{{\tan }^{\dfrac{3}{2}}}x+C\]
\[\therefore L=-2\dfrac{1}{{{\tan }^{\dfrac{1}{2}}}x}+\dfrac{2}{3}{{\left( \tan x \right)}^{\dfrac{3}{2}}}+C\]
\[\therefore L=\dfrac{-2}{\sqrt{\tan x}}+\dfrac{2}{3}{{\left( \tan x \right)}^{\dfrac{3}{2}}}+C\]
If we put the value of ‘L’ from equation (1) in the above equation we will get,
\[\therefore \int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}}dx}=\dfrac{-2}{\sqrt{\tan x}}+\dfrac{2}{3}{{\left( \tan x \right)}^{\dfrac{3}{2}}}+C\]
Therefore the value of \[\int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}}dx}\] is equal to \[\dfrac{-2}{\sqrt{\tan x}}+\dfrac{2}{3}{{\left( \tan x \right)}^{\dfrac{3}{2}}}+C\].
Therefore the correct answer is option (a).
Note: Do not use the formula \[{{\sin }^{2}}x\times {{\cos }^{2}}x=1\] in the given expression otherwise it will lead to wrong results. Also whenever you find both ‘sin’ and ‘cos’ in denominator always try to convert it in the form of ‘tan’ and ‘sec’ for simplicity.
Complete step by step answer:
To solve the given expression we will write it down first and assume it as ‘L’
\[\therefore L=\int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}}dx}\] …………………………………………………… (1)
As we see the above equation is in the form of ‘sin’ and ‘cos’ and to solve this we have to convert it in ‘tan’ so that we can solve it. If we add the powers of the term \[{{\sin }^{3}}x\times {{\cos }^{5}}x\] we will get, 3 + 5 = 8 and if we divide the term by \[{{\cos }^{8}}x\] we will get the term completely converted to ‘tan’.
To divide the term \[{{\sin }^{3}}x\times {{\cos }^{5}}x\] by \[{{\cos }^{8}}x\] we have to divide \[\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}\] by \[{{\cos }^{4}}x\] as \[\sqrt{{{\cos }^{8}}x}={{\cos }^{4}}x\].
Therefore dividing the above equation by \[{{\cos }^{4}}x\] to both the numerator and denominator we will get,
\[\therefore L=\int{\dfrac{\dfrac{1}{{{\cos }^{4}}x}}{\dfrac{\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}}{{{\cos }^{4}}x}}dx}\]
As we know that \[\dfrac{1}{{{\cos }^{4}}x}\] is equal to \[{{\sec }^{4}}x\] therefore above equation will become,
\[\therefore L=\int{\dfrac{{{\sec }^{4}}x}{\dfrac{\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}}{{{\cos }^{4}}x}}dx}\]
Also we know that \[{{\cos }^{4}}x=\sqrt{{{\cos }^{8}}x}\] therefore by replacing \[{{\cos }^{4}}x\] by \[\sqrt{{{\cos }^{8}}x}\] in the above equation we will get,
\[\therefore L=\int{\dfrac{{{\sec }^{4}}x}{\dfrac{\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}}{\sqrt{{{\cos }^{8}}x}}}dx}\]
Above equation can also be written as,
\[\therefore L=\int{\dfrac{{{\sec }^{4}}x}{\sqrt{\dfrac{{{\sin }^{3}}x\times {{\cos }^{5}}x}{{{\cos }^{8}}x}}}dx}\]
\[\therefore L=\int{\dfrac{{{\sec }^{4}}x}{\sqrt{\dfrac{{{\sin }^{3}}x\times {{\cos }^{5}}x}{{{\cos }^{\left( 3+5 \right)}}x}}}dx}\]
Now to proceed further in the solution we should know the formula given below,
Formula:
\[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\]
Referring to above formula we can replace \[{{\cos }^{\left( 3+5 \right)}}x\] by \[{{\cos }^{3}}x{{\cos }^{5}}x\] in ‘L’ therefore we will get,
\[\therefore L=\int{\dfrac{{{\sec }^{4}}x}{\sqrt{\dfrac{{{\sin }^{3}}x\times {{\cos }^{5}}x}{{{\cos }^{3}}x{{\cos }^{5}}x}}}dx}\]
As we all know that \[\dfrac{\sin x}{\cos x}=\tan x\] therefore above equation will become,
\[\therefore L=\int{\dfrac{{{\sec }^{4}}x}{\sqrt{{{\tan }^{3}}x}}dx}\]
We can also write the above equation as,
\[\therefore L=\int{\dfrac{{{\sec }^{2}}x\times {{\sec }^{2}}x}{\sqrt{{{\tan }^{3}}x}}dx}\]
As we all know the identity i.e. \[1+{{\tan }^{2}}x={{\sec }^{2}}x\] therefore above equation will become,
\[\therefore L=\int{\dfrac{\left( 1+{{\tan }^{2}}x \right)\times {{\sec }^{2}}x}{\sqrt{{{\tan }^{3}}x}}dx}\]
Now we will put, tan x = t in the above equation, ………………………………………………… (2)
Therefore, \[{{\sec }^{2}}xdx=dt\] ……………………………………………………………………………………. (3)
Also, as tan x = t
Squaring on both sides we will get,
\[{{\tan }^{2}}x={{t}^{2}}\] ………………………………………………….. (4)
If we put the values of equation (2), equation (3) and equation (4) in ‘L’ we will get,
\[\therefore L=\int{\dfrac{1+{{\operatorname{t}}^{2}}}{\sqrt{{{\operatorname{t}}^{3}}}}dt}\]
Above equation can also be written as,
\[\therefore L=\int{\dfrac{1+{{\operatorname{t}}^{2}}}{{{\operatorname{t}}^{\dfrac{3}{2}}}}dt}\]
If we separate the denominator in the above equation we will get,
\[\therefore L=\int{\left[ \dfrac{1}{{{\operatorname{t}}^{\dfrac{3}{2}}}}+\dfrac{{{\operatorname{t}}^{2}}}{{{\operatorname{t}}^{\dfrac{3}{2}}}} \right]dt}\]
\[\therefore L=\int{\left[ {{\operatorname{t}}^{-\dfrac{3}{2}}}+{{\operatorname{t}}^{2-\dfrac{3}{2}}} \right]dt}\]
\[\therefore L=\int{\left[ {{\operatorname{t}}^{-\dfrac{3}{2}}}+{{\operatorname{t}}^{\dfrac{4-3}{2}}} \right]dt}\]
\[\therefore L=\int{\left[ {{\operatorname{t}}^{-\dfrac{3}{2}}}+{{\operatorname{t}}^{\dfrac{1}{2}}} \right]dt}\]
Now to proceed further in the solution we should know the formula given below,
Formula:
\[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C\]
By using above formula in ‘L’ we will get,
\[\therefore L=\dfrac{{{\operatorname{t}}^{-\dfrac{3}{2}+1}}}{-\dfrac{3}{2}+1}+\dfrac{{{\operatorname{t}}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}+C\]
\[\therefore L=\dfrac{{{\operatorname{t}}^{1-\dfrac{3}{2}}}}{1-\dfrac{3}{2}}+\dfrac{{{\operatorname{t}}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}+C\]
\[\therefore L=\dfrac{{{\operatorname{t}}^{\dfrac{2-3}{2}}}}{\dfrac{2-3}{2}}+\dfrac{{{\operatorname{t}}^{\dfrac{1+2}{2}}}}{\dfrac{1+2}{2}}+C\]
\[\therefore L=\dfrac{{{\operatorname{t}}^{\dfrac{-1}{2}}}}{\dfrac{-1}{2}}+\dfrac{{{\operatorname{t}}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+C\]
\[\therefore L={{\operatorname{t}}^{\dfrac{-1}{2}}}\times \dfrac{-2}{1}+{{\operatorname{t}}^{\dfrac{3}{2}}}\times \dfrac{2}{3}+C\]
If we substitute the value of ‘t’ from equation (2) in the above equation we will get,
\[\therefore L=-2{{\tan }^{\dfrac{-1}{2}}}x+\dfrac{2}{3}{{\tan }^{\dfrac{3}{2}}}x+C\]
\[\therefore L=-2\dfrac{1}{{{\tan }^{\dfrac{1}{2}}}x}+\dfrac{2}{3}{{\left( \tan x \right)}^{\dfrac{3}{2}}}+C\]
\[\therefore L=\dfrac{-2}{\sqrt{\tan x}}+\dfrac{2}{3}{{\left( \tan x \right)}^{\dfrac{3}{2}}}+C\]
If we put the value of ‘L’ from equation (1) in the above equation we will get,
\[\therefore \int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}}dx}=\dfrac{-2}{\sqrt{\tan x}}+\dfrac{2}{3}{{\left( \tan x \right)}^{\dfrac{3}{2}}}+C\]
Therefore the value of \[\int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\times {{\cos }^{5}}x}}dx}\] is equal to \[\dfrac{-2}{\sqrt{\tan x}}+\dfrac{2}{3}{{\left( \tan x \right)}^{\dfrac{3}{2}}}+C\].
Therefore the correct answer is option (a).
Note: Do not use the formula \[{{\sin }^{2}}x\times {{\cos }^{2}}x=1\] in the given expression otherwise it will lead to wrong results. Also whenever you find both ‘sin’ and ‘cos’ in denominator always try to convert it in the form of ‘tan’ and ‘sec’ for simplicity.
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