Maths Integrals

${\displaystyle \underset{0}{\overset{\pi /2}{\int }}\left(\frac{1}{\left(1+ta{n}^{2}x\right)\left(1+2^x\right)}+\frac{1}{\left(1+ta{n}^{2}x\right)\left(1+2^{-x}\right)}\right)dx}$

${\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{dx}{\left(1+ta{n}^{2}x\right)}={\displaystyle \underset{0}{\overset{\pi /2}{\int }}co{s}^{2}xdx=\frac{\pi }{4}}}$

$\text{27.(32)}\begin{array}{rr}& I=2{\int }_0^{\pi /2}??\left({\mathrm{c}\mathrm{o}\mathrm{s}}^3?x+{\mathrm{c}\mathrm{o}\mathrm{s}}^4?x+{\mathrm{s}\mathrm{i}\mathrm{n}}^4?x\right)dx\\ & I=2\left[{\int }_0^{\pi /2}??{\mathrm{c}\mathrm{o}\mathrm{s}}^3?xdx+{\int }_0^{\pi /2}??{\mathrm{c}\mathrm{o}\mathrm{s}}^4?xdx+{\int }_0^{\pi /2}??{\mathrm{s}\mathrm{i}\mathrm{n}}^4?xdx\right]\\ & I=2\left[\frac{2}{3}*1+\frac{3*1}{4*2}*\frac{\pi }{2}+\frac{3*1}{4*2}*\frac{\pi }{2}\right]\\ & I=2\left[\frac{2}{3}+\frac{3\pi }{8}\right]\Rightarrow I=\frac{4}{3}+\frac{6\pi }{8}\Rightarrow 24I=32+18\pi \\ & 24I-18\pi =32\end{array}$

In the integral J, substitute x + 1 = t

$\Rightarrow dx=dtand{x}^2+2x=\left(t^2-1\right)$            

Now   $J={\displaystyle \underset{1}{\overset{e}{\int }}\frac{e^{\frac{t^2-2}{2}}}{t}}dtandK={\displaystyle \underset{1}{\overset{e}{\int }}t}lnt{e}^{\frac{t^2-2}{2}}dt$

Hence   $\left(J+K\right)={\displaystyle \underset{1}{\overset{e}{\int }}{e}^{\frac{t^2-2}{2}}}\left(\frac{1}{t}+tlnt\right)dt$

$={\left(e^{\frac{t^2-2}{2}}lnt\right)}_{t=1}^{t=e}=e^{\frac{e^2-2}{2}}={\left(\sqrt[]{e}\right)}^{e^2-2}$

$\boxed{\begin{array}{lll}\text{12.(D)}& {\int }_0^{\frac{\pi }{6}}??\frac{4{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \frac{2}{5}*2\mathrm{c}\mathrm{o}\mathrm{s}?\theta d\theta }{\left(4-4{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \right)}=\frac{16}{5*8}\int \frac{{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \mathrm{c}\mathrm{o}\mathrm{s}?d\theta }{{\mathrm{c}\mathrm{o}\mathrm{s}}^3?\theta }& x^{5/2}=2\mathrm{s}\mathrm{i}\mathrm{n}?\theta \\ & & \frac{5}{2}{x}^{3/2}dx=2\mathrm{c}\mathrm{o}\mathrm{s}?\theta d\theta \\ & & =\frac{2}{5}{\int }_0^{\frac{\pi }{6}}??{\mathrm{t}\mathrm{a}\mathrm{n}}^2?\theta d\theta =\frac{2}{5}{\int }_0^{\frac{\pi }{6}}??\left({\mathrm{s}\mathrm{e}\mathrm{c}}^2?\theta -1\right)d\theta =\frac{2}{5}[\mathrm{t}\mathrm{a}\mathrm{n}?\theta -\theta {]}_0^{\pi /6}=\frac{2}{5}\left(\frac{1}{\sqrt[]{3}}-\frac{\pi }{6}\right)\end{array}}$

$\int {\mathrm{c}\mathrm{o}\mathrm{t}}^{-1}?\left(\frac{1}{1+x}\right)dx$

$=\int {\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)dx$

$=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\int \frac{x}{1+(1+x{)}^2}dx;=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\int \frac{x}{x^2+2x+2}dx$

$=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\int \frac{\frac{1}{2}(2x+2)-1}{x^2+2x+2}dx=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\frac{1}{2}\mathrm{l}\mathrm{o}\mathrm{g}?\left(x^2+2x+2\right)+\int \frac{dx}{(x+1{)}^2+1}$

$=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\frac{1}{2}\mathrm{l}\mathrm{o}\mathrm{g}?\left(x^2+2x+2\right)+{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(x+1)+c$

Put x³/² = t
√xdx = (2/3)dt
⇒ (2/3) ∫ dt/√ (1-t²) = (2/3)sin? ¹t + c
= (2/3)sin? ¹x³/² + c
⇒ g (x) = sin? ¹x
g (0) = 0

I? = ∫? ¹ (1 − x? )? · 1 dx
= (1 - x? )? x|? ¹ - ∫? ¹ 7 (1 - x? )? (-4x³)xdx = -28∫? ¹ (1 − x? )? (1 − x? − 1)dx
I? = −28∫? ¹ (1 − x? )? dx + 28∫? ¹ (1 − x? )? dx
I? = -28I? + 28I?
29I? = 28I?
I? /I? = 28/29 ⇒ (29/4) * (I? /I? ) = (29/4) * (28/29) = 7

? C? +? C? =? C?
Σ (from r=0 to 3)? C? =? C? +? C? +? C? +? C?
=? C? +? C? +? C? +? C?
= ¹²C? = (12*11*10)/6 = 220

∫ (from 0 to π/2) (x/2)|cos (2x)|dx
∫ (from 0 to π/4) (x/2)cos (2x)dx - ∫ (from π/4 to π/2) (x/2)cos (2x)dx
= [x sin (2x)/4 - ∫sin (2x)/4 dx] (from 0 to π/4) - [x sin (2x)/4 - ∫sin (2x)/4 dx] (from π/4 to π/2)
= [x sin (2x)/4 + cos (2x)/8] (from 0 to π/4) - [x sin (2x)/4 + cos (2x)/8] (from π/4 to π/2)
= (π/16 - 1/8) - (-1/8 - π/16) = π/8

∫ (from 0 to 3) (x-1) (x-2) (x-3) dx = ∫ (from 0 to 3) (x³ - 6x² + 11x - 6) dx

A = ∫ (from 0 to 1) (x³ - 6x² + 11x - 6) dx - ∫ (from 1 to 2) (x³ - 6x² + 11x - 6) dx + ∫ (from 2 to 3) (x³ - 6x² + 11x - 6) dx
A = 11/4