I =∫0π/3esecaseca×tanada
→∫0π/3(e2−seca)(seca×tana)da
Substitute, u= e2−seca
Then,dadu=−2seca×tana→da=−seca×tana2du
On substituting, we get
∫esecaseca×tanada=−2∫eudu + C
=−2eu+C
Now put the value of u=e2−seca in above equation,
I=∣−2e−2seca∣0π/3
On solving limit, we get
I=e2−2e−1
On simplification,
I=e−1(2e−2)
Hence the value of ∫0π/3esecaseca×tanada= e−1(2e−2)
Approximation:
I=0.4773024370823822
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