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Mathematics Question on Integrals of Some Particular Functions

If ∫0π3cosx3+4sinxdx=k log(3+233)\int_{0}^{\frac{\pi}{3}}\frac{cos x}{3 + 4 sin x}dx = k\, log \left(\frac{3+2\sqrt{3}}{3}\right) then kk is

A

12\frac{1}{2}

B

13\frac{1}{3}

C

14\frac{1}{4}

D

18\frac{1}{8}

Answer

14\frac{1}{4}

Explanation

Solution

T he correct option is(C): 14\frac{1}{4}

We have,
∫0π/3cos x3+4 sin xdx\int_{0}^{\pi /3} \frac{cos\,x}{3+4\,sin\,x} dx
=k log(3+233)…(i)=k\,log \left(\frac{3+2\sqrt{3}}{3}\right)\ldots\left(i\right)
Let I=∫0π/3cos x3+4 sin xdxI=\int_{0}^{\pi/ 3} \frac{cos\,x}{3+4\,sin\,x}dx
Put 3+4sin x=t3 + 4 sin \,x = t
⇒0+4 cos x dx=dt\Rightarrow 0+4\,cos\,x\,dx =dt
Upper limit, x=π3,t=3+4 sinπ3x=\frac{\pi}{3}, t=3+4\,sin \frac{\pi}{3}
=3+4×32=3+23=3+4\times\frac{\sqrt{3}}{2}=3+2\sqrt{3}
and lower limit x=0x=0,
t=3+4 sin 0=3t=3+4\,sin \, 0=3
∴I=∫33+23dt4t=14[log t]33+23\therefore I=\int_{3}^{3+2\sqrt{3}} \frac{dt}{4t}=\frac{1}{4} \left[log\,t\right]_{3}^{3+2\sqrt{3}}
=14[log(3+23)−log 3]=\frac{1}{4}\left[log\left(3+2\sqrt{3}\right)-log\,3\right]
⇒I=14log(3+233)…(ii)\Rightarrow I=\frac{1}{4} log \left(\frac{3+2\sqrt{3}}{3}\right) \ldots\left(ii\right)
∴\therefore From Eqs. (i)\left(i\right) and (ii)\left(ii\right), we get
k=14k=\frac{1}{4}