∫x3sintan-1x41 + x8dx is equal to :
14costan-1x4 + c
14sintan-1x4 + c
- 14costan-1x4 + c
14sec-1tan-1x4 + c
In = ∫0π4tannxdx, then limn→∞nIn + In + 2 equals :
12
1
∞
zero
B.
∵ In = ∫0π4tannxdxIn + 2 = ∫0π4tann + 2xdxNow, In + In + 2 = ∫0π4tannx1 + tan2xdx = ∫0π4sec2xtannxdxLet tanx = t⇒ sec2xdx = dt∴ In + In + 2 = ∫01tndt = tn + 1n + 10 1= nn + 1Hence, limn→∞In + In + 2 = limn→∞nn + 1 = limn→∞11 + 1n = 1
If ∫xfxdx = fx2, then f(x) is equal to :
ex
e- x
log(x)
ex22
∫02x2dx is :
2 - 2
2 + 2
2 - 1
- 2 - 3 + 5
∫0πcosxdx is equal to :
- 2
- 1
∫sin2xsin3xsin5xdx is equal to :
15logesin5x - 13logesin3x + c
13logesin3x - 15logesin5x
13logesin3x + 15logesin5x
- 12cos2x + 13logesin3x
∫exlogsinx + cotxdx is equal to
excot(x) + c
exlog(sin(x)) + c
exlog(sin(x)) + tan(x) + c
ex + sin(x) + c
∫- 1010loga + xa - xdx is equal to :
0
- 2log(a + 10)
2loga + 10a - 10
2log(a + 10)
Define f(x) = ∫0xsintdt, x ≥ 0, Then :
f is increasing only in the interval 0, π2
f is decreasing in the interval 0, π
f attains maximum at x = π2
f attains minimum at x = π
Let f(x) = sin2πx1 + π2. Then, ∫fx + f- xdx is equal to :
x + c
x2 - cosπx2π + c
x2 - sin2πx4π + c