The value of ∫1 + secxdx is
sin-12sinx + C
2sin-12sinx/2 + C
2sin-12sinx + C
2sin-12x/2 + C
The value of ∫x2 + 1x4 + x2 + 1dx is
13tan-1x - 1/x3 + C
123logx - 1/x - 3x - 1/x + 3 + C
tan-1x + 1/x3 + C
tan-1x - 1/x3 + C
The value of ∫01x21 - x232dx is
132
π8
π16
π32
The value of ∫0∞x1 + xx2 + 1dx is
2π
π4
∫18 + 2x - x2dx is equal to
13sin-1x - 13 + c
sin-1x + 13 + c
13sin-1x + 13 + c
sin-1x - 13 + c
∫4ex - 252ex - 5dx = Ax + Blog2ex - 5 + c, then
A = 5 and B = 3
A = 5 and B = - 3
A = - 5 and B = 3
A = - 5 and B = - 3
B.
Let I = ∫4ex - 252ex - 5dx = ∫4ex2ex - 5dx - ∫252ex - 5dx = 4∫ex2ex - 5dx - 25∫e- x2ex - 5dxPut 2ex - 5 = u and 2 - 5e- x = v⇒ 2exdx = du⇒ 5e- xdx = dv⇒ exdx = du2 and e- xdx = dv5∴ I = 4∫du2u - 25∫du5v = 2logu - 5logv + c = 2log2ex - 5 - 5log2 - 5e- x + c = 2log2ex - 5 - 5log2ex - 5ex + c = 2log2ex - 5 - 5log2ex - 5 + 5logex + c = - 3log2ex - 5 + 5x + c⇒ I = 5x - 3log2ex - 5 + cBut it is given I = Ax + Blog2ex - 5 + c∴ A = 5 and B = - 3
∫- π2π2log2 - sinx2 + sinxdx is equal to
1
3
2
0
∫x2 + 2ax + tan-1xx2 + 1dx is equal to
loga . ax + tan-1x + c
x + tan-1xlogloga + c
ax + tan-1xloga + c
logax + tan-1x + c
If ∫fxlogsinxdx = loglogsinx + c, then f(x) is equal to
cot(x)
tan(x)
sec(x)
csc(x)
∫0π2secxnsecxn +cscxndx is equal to
π2
π3
π6