Solveeit Logo
Login

Question

Question: Evaluate \(\int\limits_0^{\dfrac{\pi }{2}} {\left( {\sqrt {\tan x} + \sqrt {\cot x} } \right)dx} \)....

Evaluate ∫0π2(tan⁡x+cot⁡x)dx\int\limits_0^{\dfrac{\pi }{2}} {\left( {\sqrt {\tan x} + \sqrt {\cot x} } \right)dx} .

Explanation

Solution

As we are given two trigonometric functions inside the integral, we know that tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{{\sin x}}{{\cos x}} and cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{{\cos x}}{{\sin x}} and now we can use the substitution method by taking sin⁡x−cos⁡x=t\sin x - \cos x = tand by changing the limits accordingly and we now the formula ∫dx1−x2=sin⁡−1x\int {\dfrac{{dx}}{{\sqrt {1 - {x^2}} }} = {{\sin }^{ - 1}}x} , using which we get the final answer.

Step by step solution :
We are asked to evaluate ∫0π2(tan⁡x+cot⁡x)dx\int\limits_0^{\dfrac{\pi }{2}} {\left( {\sqrt {\tan x} + \sqrt {\cot x} } \right)dx}
We are given two trigonometric functions inside the integral
We know that tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{{\sin x}}{{\cos x}} and cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{{\cos x}}{{\sin x}}
Using this in the given integral we get
⇒∫0π2(sin⁡xcos⁡x+cos⁡xsin⁡x)dx ⇒∫0π2(sin⁡x+cos⁡xsin⁡xcos⁡x)dx  \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left( {\sqrt {\dfrac{{\sin x}}{{\cos x}}} + \sqrt {\dfrac{{\cos x}}{{\sin x}}} } \right)dx} \\\ \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{\sin x + \cos x}}{{\sqrt {\sin x\cos x} }}} \right)dx} \\\
Let the above equation be (1)
Now let sin⁡x−cos⁡x=t\sin x - \cos x = t
Differentiating this we get (cos⁡x+sin⁡x)dx=dt\left( {\cos x + \sin x} \right)dx = dt……..(2)
Squaring t we get
⇒t2=(sin⁡x−cos⁡x)2 ⇒t2=sin⁡2x+cos⁡2x−2sin⁡xcos⁡x ⇒t2=1−2sin⁡xcos⁡x ⇒2sin⁡xcos⁡x=1−t2 ⇒sin⁡xcos⁡x=1−t22  \Rightarrow {t^2} = {\left( {\sin x - \cos x} \right)^2} \\\ \Rightarrow {t^2} = {\sin ^2}x + {\cos ^2}x - 2\sin x\cos x \\\ \Rightarrow {t^2} = 1 - 2\sin x\cos x \\\ \Rightarrow 2\sin x\cos x = 1 - {t^2} \\\ \Rightarrow \sin x\cos x = \dfrac{{1 - {t^2}}}{2} \\\
Let's use this and equation (2) in equation (1)
Now considering the limits
When x = 0 ,
⇒sin⁡0−cos⁡0=t ⇒−1=t  \Rightarrow \sin 0 - \cos 0 = t \\\ \Rightarrow - 1 = t \\\
When x=π2x = \dfrac{\pi }{2}
⇒sin⁡π2−cos⁡π2=t ⇒1=t  \Rightarrow \sin \dfrac{\pi }{2} - \cos \dfrac{\pi }{2} = t \\\ \Rightarrow 1 = t \\\
Therefore the limit ranges from – 1 to 1
⇒∫−11(dt1−t22) ⇒∫−11(2dt1−t2)  \Rightarrow \int\limits_{ - 1}^1 {\left( {\dfrac{{dt}}{{\sqrt {\dfrac{{1 - {t^2}}}{2}} }}} \right)} \\\ \Rightarrow \int\limits_{ - 1}^1 {\left( {\dfrac{{\sqrt 2 dt}}{{\sqrt {1 - {t^2}} }}} \right)} \\\
We know that ∫dx1−x2=sin⁡−1x\int {\dfrac{{dx}}{{\sqrt {1 - {x^2}} }} = {{\sin }^{ - 1}}x}
Using this we get
⇒2[sin⁡−1t]−11 ⇒2[sin⁡−1(1)−sin⁡−1(−1)] ⇒2[π2−3π2]=2(−2π2) ⇒−2π  \Rightarrow \sqrt 2 \left[ {{{\sin }^{ - 1}}t} \right]_{ - 1}^1 \\\ \Rightarrow \sqrt 2 \left[ {{{\sin }^{ - 1}}\left( 1 \right) - {{\sin }^{ - 1}}\left( { - 1} \right)} \right] \\\ \Rightarrow \sqrt 2 \left[ {\dfrac{\pi }{2} - \dfrac{{3\pi }}{2}} \right] = \sqrt 2 \left( {\dfrac{{ - 2\pi }}{2}} \right) \\\ \Rightarrow - \sqrt 2 \pi \\\

Note :
Steps to keep in mind while solving trigonometric problems are:

  1. Always start from the more complex side.
  2. Express everything into sine and cosine.
  3. Combine terms into a single fraction.
  4. Use Pythagorean identities to transform between sin⁡2x and cos⁡2x{\sin ^2}x{\text{ and }}{\cos ^2}x.
  5. Know when to apply a double angle formula.
  6. Know when to apply an additional formula.
  7. Good old expand/ factorize/ simplify/ cancelling.
  8. Integration, in mathematics, technique of finding a function g(x) the derivative of which, Dg(x), is equal to a given function f(x). This is indicated by the integral sign ∫\int {} as in ∫f(x)\int {f(x)} , usually called the indefinite integral of the function.