Integrals

I = ∫sin? ¹ (√x/√1+x)dx
∫tan? ¹ (√x)dx
= xtan? ¹√x - ∫ (1/ (1+x) * 1/ (2√x)xdx + C

= xtan? ¹√x - ∫ (t²/ (1+t²) * (t*2t dt)/ (2t) + C (x=t²)
= xtan? ¹√x - ∫ (t²/ (1+t²)dt + C = xtan? ¹√x - t + tan? ¹t + C = xtan? ¹√x - √x + tan? ¹√x + C
= (x+1)tan? ¹√x - √x + C => (Ax) = x+1 => B (x) = -√x

sinx = t, cosxdx = dt
I = ∫dt/(t³(1+t?)^(2/3)) = ∫dt/(t?(1+1/t?)^(2/3))
Put 1+1/t? = r³ ⇒ -6/t? dt = 3r²dr
I = (-1/2)∫dr = -1/2 r + c = (-1/2)(1+1/sin?x)^(1/3) + c
f(x) = (-1/2)cosec²x(1+sin?x)^(1/3) and λ=3
⇒ f(π/3) = -2

I = ∫ (cosθ / (2sin²θ + 7sinθ + 3) dθ
sinθ = t ⇒ cosθdθ = dt
= (1/2) ∫ (1 / (t² + (7/2)t + 3/2) dt
= (1/2) ∫ (1 / (t+7/4)² - (5/4)²) dt
= (1/5) ln | (2t+1)/ (t+3)| + c
= (1/5) ln | (2sinθ+1)/ (sinθ+3)| + C
So A = 1/5
B (θ) = 5 (2sinθ+1)/ (sinθ+3)

$?{\mathcal{l}}_{1}and{\mathcal{l}}_2$ are perpendicular, so

$3*1+\left(-2\right)\left(\frac{\alpha }{2}\right)+0*2=0$

a = 3

Now angle between ${\mathcal{l}}_{2}and{\mathcal{l}}_3$ ,

$cos\theta =\frac{1\left(-3\right)+\frac{\alpha }{2}\left(-2\right)+2\left(4\right)}{\sqrt[]{1+\frac{{\alpha }^2}{4}+4}.\sqrt[]{9+4+16}}$

$\Rightarrow cos\theta =\frac{\frac{2}{29}}{2}\Rightarrow \theta =co{s}^{-1}\left(\frac{4}{29}\right)=se{c}^{-1}\left(\frac{29}{4}\right)$

$\underset{x\to \frac{1}{\sqrt[]{2}}}{lim}\frac{sin\left(co{s}^{-1}x\right)-x}{1-tan\left(co{s}^{-1}x\right)}$

let $co{s}^{-1}x=\frac{\pi }{4}+\theta$

= $\underset{\theta \to \infty }{lim}\frac{\text{exception 25:}sin\theta }{-2tan\theta }\left(1-tan\theta \right)=-$$\frac{1}{}$