I I I = ∫ 0 π / 3 s e c a × t a n a e s e c a d a =\int_{0}^{\pi/3} \dfrac{seca\times tana}{\sqrt{e^{seca}}}da = ∫ 0 π /3 e sec a sec a × t ana d a
→ ∫ 0 π / 3 ( e − s e c a 2 ) ( s e c a × t a n a ) d a \rightarrow \int_{0}^{\pi/3} (e^{\dfrac{-seca}{2}})(seca\times tana )da → ∫ 0 π /3 ( e 2 − sec a ) ( sec a × t ana ) d a
Substitute, u= e − s e c a 2 e^{\dfrac{-seca}{2}} e 2 − sec a
T h e n , d u d a = − s e c a × t a n a 2 → d a = − 2 s e c a × t a n a d u Then, \dfrac{du}{da}=- \dfrac{seca\times tana}{2} \rightarrow da=- \dfrac{2}{seca\times tana}du T h e n , d a d u = − 2 sec a × t ana → d a = − sec a × t ana 2 d u
On substituting, we get
∫ s e c a × t a n a e s e c a d a = − 2 ∫ e u d u \int\dfrac{seca\times tana}{\sqrt{e^{seca}}}da=-2 \int e^udu ∫ e sec a sec a × t ana d a = − 2 ∫ e u d u + C
= − 2 e u + C = -2 e^u +C = − 2 e u + C
Now put the value of u=e − s e c a 2 e^{\dfrac{-seca}{2}} e 2 − sec a in above equation,
I = ∣ − 2 e − s e c a 2 ∣ 0 π / 3 I= \mid-2e^{-\dfrac{seca}{2}}\mid_{0}^{\pi/3} I =∣ − 2 e − 2 sec a ∣ 0 π /3
On solving limit, we get
I = 2 e − 2 e − 1 I=\dfrac{2}{\sqrt e} - 2e^{-1} I = e 2 − 2 e − 1
On simplification,
I = e − 1 ( 2 e − 2 ) I=e^{-1}(2\sqrt e-2) I = e − 1 ( 2 e − 2 )
Hence the value of ∫ 0 π / 3 s e c a × t a n a e s e c a d a = \int_{0}^{\pi/3} \dfrac{seca\times tana}{\sqrt{e^{seca}}}da= ∫ 0 π /3 e sec a sec a × t ana d a = e − 1 ( 2 e − 2 ) e^{-1}(2\sqrt e-2) e − 1 ( 2 e − 2 )
Approximation:
I = 0.4773024370823822 I=0.4773024370823822 I = 0.4773024370823822
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