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please integrate this definite integration at picture bilow

please integrate this definite integration at picture bilow

Question Image
Grade:

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int_{0}^{\pi /2}\frac{1}{5+4sinx}dx
I = \int_{0}^{\pi /2}\frac{1}{5+4sin(\frac{\pi }{2}-x)}dx
I = \int_{0}^{\pi /2}\frac{1}{5+4cosx}dx
I = \int_{0}^{\pi /2}\frac{1}{5+4(2cos^{2}\frac{x}{2}-1)}dx
I = \int_{0}^{\pi /2}\frac{1}{8cos^{2}\frac{x}{2}+1}dx
I = \int_{0}^{\pi /2}\frac{sec^{2}\frac{x}{2}}{sec^{2}\frac{x}{2}+8}dx
I = \int_{0}^{\pi /2}\frac{sec^{2}\frac{x}{2}}{tan^{2}\frac{x}{2}+1+8}dx
I = \int_{0}^{\pi /2}\frac{sec^{2}\frac{x}{2}}{tan^{2}\frac{x}{2}+9}dx
t = tan\frac{x}{2}
dt = sec^{2}\frac{x}{2}.\frac{1}{2}dx
x=0\rightarrow t=0
x=\frac{\pi }{2}\rightarrow t=1
I = \int_{0}^{1}\frac{2}{t^{2}+9}dt
I = \int_{0}^{1}\frac{2}{t^{2}+(3)^{2}}dt
I = \frac{2}{3}[tan^{-1}(\frac{t}{3})]_{0}^{1}
I = \frac{2}{3}tan^{-1}(\frac{1}{3})

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