BITSAT 10.........

(#) ∫(0 to pi/2) log sinx dx = ?

(A) pi/2 log 2

(B) - pi/2 log 2

Ans given B

($) If the lines y = m_i x, i = 1,2,3. cut equal intercepts on the transversal line x + y = 1 , then

1+m1 , 1+ m2 , 1+m3 are in

(A) AP

(b) GP

(C) HP

(D) None

Ans given C i got B,C

(1) Newspaper Hoarding and Television Broadcast.

(2) Material growth's meanings are vague here. I suppose it would be internal, but it rather means "worldly objects and desires".

(3) This is an NCERT integration question. Ans is -pi/2 log2.

Q2)
I = \int_{0}^{\pi /2}{logsinxdx};also ;I=\int_{0}^{\pi /2}{logcosxdx}
adding these
2I = \int_{0}^{\pi /2}{(logsinx+logcosx)dx}=\int_{0}^{\pi /2}{log\left(\frac{sin2x}{2} dx\right)}

= \int_{0}^{\pi /2}{logsin2xdx}-\int_{0}^{\pi /2}{log2}dx

Let I₁ = \int_{0}^{\pi /2}{logsin2xdx}
put 2x = t → dx = dt2

\frac{1}{2}\int_{0}^{\pi}{logsintdt}
\frac{1}{2}\int_{0}^{2.\frac{\pi}{2}}{logsinxdx} [ just replaced t by x ]
\frac{2}{2}\int_{0}^{\frac{\pi}{2}}{logsinxdx}=I

hence we got I₁ = I

now we had 2I = I₁ -\int_{0}^{\frac{\pi}{2}}{log2}
2I = I -\int_{0}^{\frac{\pi}{2}}{log2}
so I = -\int_{0}^{\frac{\pi}{2}}{log2}

@Pritish : Y is the meaning vague

n how do u say internal I dont get u ??

@Manmay : I too have this solution ya (given in the solution)

But i wanted a shorter method (if any ) coz this is a qs from BITS exam na thats y

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