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Question: Evaluate the following definite integral: \(\int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left( {\df...

Evaluate the following definite integral: ∫−π4π4(11+sin⁡x)dx\int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left( {\dfrac{1}{{1 + \sin x}}} \right)dx} r.

Explanation

Solution

Hint- Rationalize the given function which needs to be integrated.

Let the given integral be I=∫−π4π4(11+sin⁡x)dxI = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left( {\dfrac{1}{{1 + \sin x}}} \right)dx}
Now rationalizing the function on the RHS of the above equation, we get
I=∫−π4π4[(11+sin⁡x)×(1−sin⁡x1−sin⁡x)]dx=∫−π4π4[1−sin⁡x(1+sin⁡x)(1−sin⁡x)]dx ⇒I=∫−π4π4[1−sin⁡x1−(sin⁡x)2]dx  I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\left( {\dfrac{1}{{1 + \sin x}}} \right) \times \left( {\dfrac{{1 - \sin x}}{{1 - \sin x}}} \right)} \right]dx} = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\dfrac{{1 - \sin x}}{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}} \right]dx} \\\ \Rightarrow I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\dfrac{{1 - \sin x}}{{1 - {{\left( {\sin x} \right)}^2}}}} \right]dx} \\\
Using the identity (sin⁡x)2+(cos⁡x)2=1⇒1−(sin⁡x)2=(cos⁡x)2{\left( {\sin x} \right)^2} + {\left( {\cos x} \right)^2} = 1 \Rightarrow 1 - {\left( {\sin x} \right)^2} = {\left( {\cos x} \right)^2} , we have
⇒I=∫−π4π4[1−sin⁡x(cos⁡x)2]dx=∫−π4π4[1(cos⁡x)2−sin⁡x(cos⁡x)(cos⁡x)]dx\Rightarrow I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\dfrac{{1 - \sin x}}{{{{\left( {\cos x} \right)}^2}}}} \right]dx = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\dfrac{1}{{{{\left( {\cos x} \right)}^2}}} - \dfrac{{\sin x}}{{\left( {\cos x} \right)\left( {\cos x} \right)}}} \right]dx} }
Since, sec⁡x=1cos⁡x\sec x = \dfrac{1}{{\cos x}} and tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{{\sin x}}{{\cos x}}

∴I=∫−π4π4[(sec⁡x)2−(tan⁡x)(sec⁡x)]dx=∫−π4π4(sec⁡x)2dx−∫−π4π4(tan⁡x)(sec⁡x)dx ⇒I=[tan⁡x]−π4π4−[sec⁡x]−π4π4=[tan⁡(π4)−tan⁡(−π4)]−[sec⁡(π4)−sec⁡(−π4)] ⇒I=[tan⁡(π4)−tan⁡(−π4)]−[sec⁡(π4)−sec⁡(−π4)]  \therefore I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {{{\left( {\sec x} \right)}^2} - \left( {\tan x} \right)\left( {\sec x} \right)} \right]dx} = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {{{\left( {\sec x} \right)}^2}dx - \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left( {\tan x} \right)\left( {\sec x} \right)} dx} \\\ \Rightarrow I = \left[ {\tan x} \right]_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} - \left[ {\sec x} \right]_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} = \left[ {\tan \left( {\dfrac{\pi }{4}} \right) - \tan \left( { - \dfrac{\pi }{4}} \right)} \right] - \left[ {\sec \left( {\dfrac{\pi }{4}} \right) - \sec \left( { - \dfrac{\pi }{4}} \right)} \right] \\\ \Rightarrow I = \left[ {\tan \left( {\dfrac{\pi }{4}} \right) - \tan \left( { - \dfrac{\pi }{4}} \right)} \right] - \left[ {\sec \left( {\dfrac{\pi }{4}} \right) - \sec \left( { - \dfrac{\pi }{4}} \right)} \right] \\\

Since, we know that tan⁡(−θ)=tan⁡θ\tan \left( { - \theta } \right) = \tan \theta and sec⁡(−θ)=sec⁡θ\sec \left( { - \theta } \right) = \sec \theta
∴I=[tan⁡(π4)+tan⁡(π4)]−[sec⁡(π4)−sec⁡(π4)]=2tan⁡(π4)\therefore I = \left[ {\tan \left( {\dfrac{\pi }{4}} \right) + \tan \left( {\dfrac{\pi }{4}} \right)} \right] - \left[ {\sec \left( {\dfrac{\pi }{4}} \right) - \sec \left( {\dfrac{\pi }{4}} \right)} \right] = 2\tan \left( {\dfrac{\pi }{4}} \right)
Also, tan⁡(π4)=1\tan \left( {\dfrac{\pi }{4}} \right) = 1
⇒I=2×1=2\Rightarrow I = 2 \times 1 = 2.

Note- These types of problems can be solved by rationalizing the function which needs to be integrated in order to get a function for which the formula of integration is known.