Solution:

We have, `a+ b = 0 => a = - b`

`:. int_a^b (x^7 + sin x)/(cos x) dx = int_(-b)^b (x^7/(cos x) + (sin x)/(cos x)) dx`

` = int_(-b)^b (x^7 . sec x + tan x) dx`

The given integrand is an odd function.

Hence, its answer is `0`.

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Solution:

Let `I = int _(-pi//2)^(pi//2) x sin x dx`

` I = 2 int _(0)^(pi//2) x sin x dx`

`[∵ f(x) = x sin x` is an even function.

` => f(- x) = (- x) sin(- x) = (- x) (-sin x) = x sin x = f(x)]`

` [∵ int_(-a)^(a) f(x) = {2 int_(-a)^(a) f(x)`, if `f(x)` is an even function

` = 0`, if `f(x)` is an odd function]

---

Solution:

Let `f(x) = sin^3 x cos^2 x`

`f(-x) =sin^3 (-x) cos^2 (-x)`

`=-sin^3 x cos^2 x=- f(x)`

`:.` `f(x)` is an odd function.

`:. I= 0`

---

Solution:

Given,

`int_(-2)^5 f(x) dx = 4, int_0^5 {1 + f(x)} dx = 7`

Let `I = int_(-2)^5 f(x) dx = int_(-2)^0 f(x) dx + int_5^0 f(x) dx`

`=> 4 = int_(-2)^0 f(x) dx + int_0^5 [(1 + f(x)) - 1] dx`

`=> 4 = int_(-2)^0 f(x) dx + int_0^5 [1 + f(x)] dx - int_0^5 1 dx`

` => 4 = int_(-2)^0 f(x) dx + 7 - (x)_0^5`

`=> 4 = int_(-2)^0 f(x) dx + 7 - 5`

`=> 4 = int_(-2)^0 f (x) dx + 2`

`=> int_(-2)^0 f(x) dx = 2`

---

Solution:

Let `I = A = int _0^pi (sin x dx)/(sin x + cos x) ` ..........(i)

Let `I = B = int _0^pi (sin x dx)/(sin x - cos x) ` .........(ii)

` [∵ int_0^a fx(x)dx = int_0^a f(a - x) dx]`

On adding Eqs. (i) and (ii), we get

` 2I = int_0^(pi) ( (sin x)/( sin x + cos x) + (sin x)/( sin x - cos x) ) dx`

`=> 2I = int_0^(pi) ( sin x (sin x - cos x + sin x + cos x))/( sin^2 x- cos^2 x) dx`

`=> 2I = 4 int_0^(pi//2) (sin^2 x)/(sin^2 x- cos^2 x) ` .........(iii)

` [ ∵ int_0^(2a) f(x) dx = 2 int_0^a f(x) dx` is ` f ( 2a - x) = f(x) ]`

` => 2I = 4 int_0^(pi//2) (cos^2 x)/(cos^2 x- sin^2 x) ` .........(iv)

`[ ∵ int_0^a f(x) dx = int_0^a f( a - x) dx ]`

` => 4I = 4 int_0^(pi//2) ( ( sin^2 x - cos^2 x)/( sin^2 x - cos^2 x)) dx`

[On adding Eqs. (iii) and (iv)]

` => 4I = 4(x)_0^(pi//2) `

`=> 4I = 4 xx pi/2 => I = pi/2`

---

Solution:

We know that

`int_(-3)^9f(x)dx = int_(-3)^2f(x)dx+int_2^9f(x)dx` (by property) ... (i)

given `int_(-3)^(2) f(x)dx = 7/3` and `int_(-3)^9f(x)dx = -5/6`

`[becauseint_a^bf(x)dx = int_a^cf(x)dx+int_c^bf(x)` where `alecleb]`

from eq(i)


`-5/6 = 7/3+int_2^9f(x)dx`

`int_2^9f(x)dx = -5/6-7/3 = (-5-14)/6 = -19/6`

---

Solution:

Given, `I = int_0^pi ( x dx)/(1 +sinx)` .......(1)

` = int_0^pi (pi - x)/( 1 + sin (pi - x)) dx`

`[∵ int_0^a f(x) dx = int_0^a f(a- x)dx ]`

` = int_0^pi (pi - x)/(1 + sin x) dx` ..........(2)

`[∵ sin (pi - x) = sin x]`

On adding Egs. (1) and (2), we get

` 2I = pi int_0^(pi) (dx)/(1 + sin x)` ..........(3)

` => 2I = 2pi int_0^(pi/2) (dx)/(1 + sin x)`

`[ ∵ int_0^(2a) f(x) dx = 2 int_0^a f(x) dx ,` if `f(2a - x) = f(x)]`

` => I = pi int_0^(pi/2) (dx)/((1 + (2 tan (x/2) ))/( 1 = tan^2 (x/2) ))`

` => I = pi int_0^(pi/2) ( sec ^2 (x/2) dx)/( tan^2 (x/2) + 1 + 2 tan (x/2))`

` => I = pi int_0^(pi/2) (( sec^2 (x/2) )dx)/( tan (x/2) + 1)`

Put `tan (x/2) + 1 = t`

`=> sec^2 (x/2) . 1/2 dx = dt`

`=> sec^2 (x/2) dx = 2 dt`

When `x = 0` then `t = 1` and when `x = pi/2 `. then `t = 2`

`:. I = 2pi int_1^2 ( dt)/t^2 = -2pi [1/t]_1^2= -2pi [1/2 -1]`

` = -2 pi (-1/2 ) = pi`

According to the explanation, `I = pi`

---

Solution:

Let `I = int_0^1 underset(I)x underset(II)e^x dx`

Using integration by parts,

`I = [x*e^x]_0^1-int_0^1 1*e^xdx`


`=> I = [1*e -0]-[e^x]_0^1 => I = e-(e-1) = e-e+1`

`therefore I = 1`

---

Solution:

`I = int_0^pi (dx)/(1+2sin^2x) = 2 int_0^(pi/2) (dx)/(1+2sin^2x)` `[because int_0^(2a) f(x)dx = 2int_0^a f(x)dx]`

If `{f(2a-x) = f(x)}`


`I = 2int_0^(pi/2) (dx)/(1+1-cos2x) = 2int_0^(pi/2)(dx)/(2-cos2x)`


` = 2int_0^(pi/2) (dx)/(2-((1-tan^2x)/(1+tan^2x))) = 2int_0^(pi/2) (sec^2x)/(1+3tan^2x)dx`

Let `sqrt3tanx = t => sqrt3sec^2xdx = dt`


`I = 2 int_0^(oo) (dt)/(sqrt3 (1+t^2)) = 2/sqrt3[tan^(-1)t]_0^(oo)`


` = 2/sqrt3[tan^(-1)(oo)-tan^(-1)(0)]`

` = 2/sqrt3(pi/2-0) = (2pi)/(2sqrt3) = pi/sqrt3`

---

Solution:

`because f(pi-x) = sin^7 (pi-x)`

` = sin^7 x = f(x)`

`therefore int_0^(pi) sin^7 xdx = int_0^((2pi)/2) sin^7 x dx`


` = 2int_0^(pi/2) sin^7xdx`


Also `sin^7 x` is an odd function `[because f(-x) = f(x)]`

Hence, both A and Rare individually true but R is not the correct
explanation of A.

---

Solution:

Let ` I = int_0^(pi//2) ( sin^2 x)/(sin x + cos x ) dx ` .........(1)

` => I = int_0^(pi//2) ( sin^2 (pi/2 - x) ) /(sin (pi/2 - x) + cos ( pi/2 - x) ) dx`

` =>I = int_0^(pi//2) ( cos^2 x)/(sin x + cos x ) dx` ........(ii)

Adding,(i) and (ii),

` => 2 I = int_0^(pi//2) ( sin^2 x + cos ^2 x )/(sin x + cos x ) dx`

` => 2 I = int_0^(pi//2) (dx)/(sin x + cos x ) `

` => 2 I = 1/sqrt2 int_0^(pi//2) (dx)/(sin x . 1/sqrt2 + cos x 1/sqrt2 ) `

` => 2 I = 1/sqrt2 int_0^(pi//2) (dx)/(sin x . cos pi/4 + cos x . sin pi/4 ) `

` => 2 I = 1/sqrt2 int_0^(pi//2) (dx)/(sin (pi/4 + x ))`

` => 2 I = 1/sqrt2 int_0^(pi//2) cosec ( pi/4 + x ) dx`

` => 2 I = 1/sqrt2 [ I n | cosec ( pi/4 + x ) - cot ( pi/4 + x ) | ]_0^(pi//2)`

` => 2 I = 1/sqrt2 [ I n | cosec ( pi/4 + pi/2 ) - cot ( pi/4 + pi/2) | - I n | cosec ( pi/4 + 0 ) - cot ( pi/4 + 0 ) | ]`

` => 2 I = 1/sqrt2 [ I n | sqrt2 - (-1) | - I n | sqrt2 - 1 | ]`

` => I = 1/sqrt2 [ I n | ( sqrt2 + 1)/(sqrt 2 - 1) | ]`

---

Solution:

Let `I = int _1^x (logx^2)/x dx`

`= int _1^x (2 log x) /x dx`

Put, `log x =t => 1/x dx = dt`

`:. I =2 int_0^(logx) t dt = [t^2]_0^(logx)`

`= (logx)^2`

---

Solution:

Given, `f(x) = { tt ((2x^2+1, x le 1),(4x^3-1, x > 1))`

`:. int_0^2 f(x) =int_0^1 f(x) dx +int_1^2 f(x) dx`

`=int_0^1 (2x^2+1) dx +int_1^2 (4x^3-1) dx`

`=[(2x^3)/3+x]_0^1 +[(4x^4)/4 -x]_1^2`

`=2/3 (1)^3 +1-(0+0) +{(2)^4-2-{(1)^4-1}]`

`=2/3 +1 +[16-2-0]`

`=2/3 +15 =(2+45)/3`

`=47/3 ` sq units

---

Solution:

Let `I = int_0^1 ( log (1 + x))/(1 + x^2) dx`

On putting `x = tan theta`, we get

` I = int_0^(pi//4) log ( 1 + tan theta) d theta` ..........(i)

` => I = int_0^(pi//4) log { 1 + tan ( pi/4 - x) }`

` => I = int_0^(pi//4) log ( 1 + (1 - tan x)/(1 + tan x) ) dx `

` => I = int_0^(pi//4) log ( 2/(1 + tan x) ) dx` .......(ii)

On adding Eqs. (i) and (ii), we get

` 2I = int_0^(pi//4) log 2 dx = pi/4 log_e 2`

` => I = pi/8 log_e 2`

---

Solution:

Let `I = int_0^(pi//2) (sin^2 x)/(sin x +cos x) dx` ......(i)

Now, ` I = int_0^(pi//2) ( sin^2 ( pi/2 - x))/( sin ( pi/2 - x) + cos ( pi/2 - x) )`

` I = int_0^(pi//2) (cos^2 x)/(sin x +cos x) dx` .....(ii)

On adding Eqs. (i) and (ii), we get

` 2I = int_0^(pi//2) ( sin^2 x + cos^2x)/(sin x + cos x) dx`

` = int_0^(pi//2) 1/(sin x + cos x) dx`

` = 1/sqrt(2) int_0^(pi//2) 1/( cos ( x - pi/4)) dx`

` = 1/sqrt(2) int_0^(pi//2) sec ( x - pi/4) dx`

` = 1/sqrt(2) [ log { sec ( x - pi/4 ) + tan ( x - pi/4) } ]_0^(pi//2)`

` = 1/sqrt(2) [ log ( sqrt2 + 1) - log ( sqrt2 - 1)]`

` = 1/sqrt(2) log ( ( sqrt2 + 1)/(sqrt2 - 1) )`

` = 1/sqrt(2) log ( sqrt2 + 1)^2`

` = sqrt(2) log ( sqrt2 + 1)`

---

Solution:

We have `int_0^pi x f(sin^2 x+sec^x )dx`

`=k int_0^(pi//2) f(sin^2 x+sec^2 x)dx`

Let `I=int_0^pi x f(sin^2 x+sec^2 x)dx`...............(i)

`=int_0^pi (pi -x) f(sin^2 (pi-x) +sec^2(pi -x)) dx`

`=int_0^pi (pi-x) f(sin^2 x+sec^2 x)dx`.............(ii)

On addinq Eqs.(1) and (ii), we qet

`2I=pi int_0^pi f(sin^2 x+sec^2 x) dx`

`=> 2I=2 pi int_0^(pi//2) f(sin^2 x+sec^2 x) dx`

`=> I=pi int_0^(pi//2) f(sin^2 x+sec^2 x) dx`

On comparing with given integral. we qet `k = pi`

---

Solution:

`I_1 = 3 int_(0)^(pi) f(cos^2 x)dx = 3I_2` `[because text(period is) pi]`

---

Solution:

Let `I = int _0^1 (e^(tan^(-1)quad_x) dx)/(1 + x^2)`

put ` t = tan^(-1) x `

` dt = 1/(1 + x^2) dx`

Lower Limit `-> t = tan^(-1) 0 = 0`

Upper Limit `-> t = tan^(-1) 1 = pi/4`

` :. I = int _0^(pi/4) e^t = [e^t] _0^(pi/4)`

` = [e^(pi/4) - e^0 ] = e^(pi/4) -1`

---

Solution:

Let `I = int_0^(pi/2) (dx)/(a^2 cos^2 x + b^2 sin^2 x)`

` = int_0^(pi/2) (sec^2 x dx)/(a^2 + b^2 tan^2 x)`

[divide numerator and denominator by `cos^2 x`]

Put `tan x = t => sec^2 x dx = dt`

When `x = 0`, then `t = 0` and when `x = pi/2` then `t = oo`

`:. I = int_0^(oo) (dt)/(a^2 + b^2t^2) = 1/b^2 int_0^(oo) (dt)/((a/b)^2 + t^2)`

` = 1/b^2 1/(a/b) [ tan^(-1) ((bt)/a) ]_0^(oo)`

` [ ∵ int (dx)/(a^2 + x^2) = 1/a tan^(-1) (x/a) + C]`

` 1/(ab) [tan^(-1)(oo) - tan^(-1)(0)]`

` 1/(ab) [pi/2 - 0] = pi/(2ab)`

---

Solution:

We note that `x^3 – x ≥ 0` on `[– 1, 0]` and `x^3 – x ≤ 0` on `[0, 1]` and that `x^3 – x ≥ 0` on `[1, 2]`. So by `P_2` we write

`int_(-1)^2 | x^3 -x| dx = int_(-1)^0 (x^3 -x) dx+ int_0^1 -(x^3 -x) dx + int_1^2 (x^3 -x) dx`

`=int_(-1)^0 (x^3 -x) dx + int_0^1 (x -x^3) dx + int_1^2 (x^3 -x) dx`

`=[ x^4/4 - x^2/2]_(-1)^0 +[ x^2/2 - x^4/4]_0^1 +[ x^4/4 - x^2/2]_1^2`

`=- (1/4 -1/2)+ (1/2 - 1/4) +(4 -2) -(1/4 - 1/2)`

`=1/4 + 1/2 + 1/2 - 1/4 + 2 - 1/4 + 1/2 =3/2 - 3/4 + 2 = 11/4`

---

Solution:

` I = int_(-5)^(-2) | x + 2 | dx + int_(-2)^5 | x + 2 | dx`

` = - int_(-5)^(-2) x + 2 dx + int_(-2)^5 x + 2 dx`

` = - [ x^2/2 + 2 x]_(-5)^(-2) + [ x^2/2 + 2x ]_(-2)^5`

` = - [ (4/2 - 4) - ( (25)/2 - 10) ] + [ ( (25)/2 + 10) - (4/2 - 4) ]`

` = 29`

---

Solution:

`I = int_(e^(-1))^(e^2) |(lnx)/x|` dx

As both `e^(-1)` and `e^2` are positive `| x | = x`

Let `ln x = t`

`1/x dx = dt`

`I = int_(-1)^(2) |t| dt`

`= (int_(-1)^(0) -t + int_(0)^(2) t ) dt`

`= -[t^2 /2]_(-1)^0 + [t^2 /2]_(0)^2`

`= 3/2`

---

Solution:

Let ` I = int _0^(4 pi) | cos x | dx = 2 int _0^(2 pi) | cos x | dx`

`[because int_0^(2a) f(x) dx = 2 int _0^a f(x) dx`, if `f(2a -x) = f(x) `, here

`I cos (4pi- x) I= I cos xI]`

`2 * 2 int_0^(2a) f(x) dx = 2 int_0^a | cos x |dx ` `[ because | cos(2 pi -x)| = | cos x|]`

`= 2 * 2 * 2 int _0^(pi//2) | cos x | dx` `\ \ \ \ [ | cos ( pi -x ) | = | - cos x | = | cos | ]`

`=8 int _0^(pi/2) cos x dx ` `[ because cos x > 0 , AA x in (0, pi/2)]`

`= 8 [sin ] _0^(pi/2) = 8 [ sin frac (pi)(2) - sin 0 ] =8`

---

Solution:

The capsule is the protective covering of the bacterial cell made up of polymers of simple sugars (polysaccharides). Most capsules are hydrophilic and may help the bacterium avoid desiccation. It protects a bacterial cell from ingestion and destruction by white blood cells (phagocytosis) and helps it to hide from host immune system.

---

Solution:

` int _(-1)^1 x |x| dx = int _(-1)^0 x (-x) dx + int _(0)^1 x (x) dx`

[when `a < c < b`, then ` int _(a)^b f( x) dx = int _(a)^c f( x) dx + int _(c)^b f( x) dx `]

` = int _(-1)^0 - x^2 dx + int _(0)^1 x^2 dx`

` = [- x^3/3]_(-1)^0 + [ x^3/3]_(0)^1`

` = 0 - (-1/3) + 1/3 = 2/3 `

---

Solution:

` int _a^b (|x|)/x dx = int _a^b 1 dx quad [ ∵ 0 < a < b ]`

`= [x] _a^b = | b|- |a|`

---

Solution:

Given, `g(x) = [1/x]`

When `1/3 < x < 1/2 => 2 < 1/x < 3`

`:. [1/x] = 2`

Now, we can find ` int _(1//3)^(1//2) g(x) dx`

`:. int _(1//3)^(1//2) g(x) dx = int _(1//3)^(1//2) [1/x] dx = int _(1//3)^(1//2) 2dx`


` = 2[x]_(1//3)^(1//2) =2 [1/2 -1/3]`

` = 2 xx (3-2)/6 = 2 xx 1/6 = 1/3 `

---

Solution:

Given, `f(x) = x g(x)`

and `g(x)= [1/x]`

We have to split given limit into two parts.

When `1/3 < x < 1/2 => 2 < 1/x < 3`

`=> [1/x ] = 2`

`f(x) = x g(x) = 2x`

When `1/2 < x < 1 => 1 < 1/x < 2 => [1/x] = 1`

`:. f(x) = x g(x) = x`

Now ` int _(1//3)^(1) f(x) dx = int _(1//3)^(1//2) 2x dx + int_(1//2)^1 x dx`

` = 2 [x^(2)/2]_(1//3)^(1//2) + [x^(2)/2]_(1//2)^(1)`

` = 2/2 [(1/2)^(2) - (1/3)^(2)] +1/2 [(1)^(2) - (1/2)^(2)]`

` = [1/4 - 1/9] + 1/2 [1- 1/4]`

` = (9-4)/36 + 1/2 [(4-1)/4] = 5/ (36) = 3/8 = (10 +27)/72 = 37/72`

---

Solution:

`int _(-2)^(2) x dx - int _(-2)^(2) [x] dx`

We have, `[x] = {tt ((-2 , -2 <= x < -1) ,(-1 , -1 <= x < 0) , (0 , 0 <= x < 1), (1, 1 <= x < 2 ))`

` :. int _(-2)^(2) x dx - [int _(-2)^(-1) (-2) dx + int _(-1)^(0) (-1) dx + int _(0)^(1) 0 dx + int _(1)^(2) 1 dx ]`

` = 0 - [ -2 (x) _(-2)^(-1) + (-x) _(-1)^(0) + (x)_(1)^(2)]`

`[ ∵ int _(-a)^(a) f(x) dx = 0 ,` if ` f(x)` is odd`]`

`=-[-2(-1+ 2)+ (0- 1)+ (2- 1)]`

`=-[-2-1+1]=2`

---

Solution:

Let `I= int_(0)^(pi) sin^(50) x cos^(49) x dx`

`= int_(0)^(pi) sin^(50) (pi-x) cos^(49) (pi- x) dx`

`= - int_(0)^(pi) sin^(50) x cos^(49) x dx`

`=> I = -I => I =0`

---

Solution:

Let `I= int_(0)^(pi/2) (sqrt (sec x) )/(sqrt (cosec x) + sqrt (sec x)) dx`............(i)

`=> I= int_(0)^(pi/2)`

`(sqrt (sec (pi/2 -x) ) )/(sqrt (cosec (pi/2 -x) ) sqrt (sec (pi/2 -x) )) dx`

`=> I = int_(0)^(pi/2) (sqrt (cosec x) )/(sqrt (sec x) + sqrt (cosec x)) dx`...........(ii)

On adding Eqs. (i) and (ii), we get

`2I = int_(0)^(pi/2) dx = [x]_(0)^(pi/2)`

`=> I = pi/4`

---

Solution:

Let ` I = int_0^(pi//2) ( sin^2 x)/(sin x + cos x ) dx ` .........(1)

` => I = int_0^(pi//2) ( sin^2 (pi/2 - x) ) /(sin (pi/2 - x) + cos ( pi/2 - x) ) dx`

` =>I = int_0^(pi//2) ( cos^2 x)/(sin x + cos x ) dx` ........(ii)

Adding,(i) and (ii),

` => 2 I = int_0^(pi//2) ( sin^2 x + cos ^2 x )/(sin x + cos x ) dx`

` => 2 I = int_0^(pi//2) (dx)/(sin x + cos x ) `

` => 2 I = 1/sqrt2 int_0^(pi//2) (dx)/(sin x . 1/sqrt2 + cos x 1/sqrt2 ) `

` => 2 I = 1/sqrt2 int_0^(pi//2) (dx)/(sin x . cos pi/4 + cos x . sin pi/4 ) `

` => 2 I = 1/sqrt2 int_0^(pi//2) (dx)/(sin (pi/4 + x ))`

` => 2 I = 1/sqrt2 int_0^(pi//2) cosec ( pi/4 + x ) dx`

` => 2 I = 1/sqrt2 [ I n | cosec ( pi/4 + x ) - cot ( pi/4 + x ) | ]_0^(pi//2)`

` => 2 I = 1/sqrt2 [ I n | cosec ( pi/4 + pi/2 ) - cot ( pi/4 + pi/2) | - I n | cosec ( pi/4 + 0 ) - cot ( pi/4 + 0 ) | ]`

` => 2 I = 1/sqrt2 [ I n | sqrt2 - (-1) | - I n | sqrt2 - 1 | ]`

` => I = 1/sqrt2 [ I n | ( sqrt2 + 1)/(sqrt 2 - 1) | ]`

---

Solution:

Since, period of `| sin x | ` is `pi`.

`:. I = int_(pi)^(16 pi) | sin x | dx`

`= 15 int_(0)^(pi) sin x dx = 15 [-cos x]_(0)^(pi)`

`= 15 (- cos pi + cos 0) =15 (1+1)`

`30`

---

Solution:

`I = int_0^(pi/2) sin^0 x * cos^8 xdx `

`(because int_0^(pi/2) sin^m x * cos^n xdx = (r((m+1)/2)r((n+1)/2))/(2sqrt((m+n+2)/2)))`



`I = (r((0+1)/2)r((8+1)/2))/(2r((0+8+2)/2))`


` = (r(1/2) r (9/2))/(2r 5) ` `(becausern = nrn)`


` = sqrtpi * (7/2 * 5/2 * 3/2 * 1/2 * sqrtpi )/(2 * 4 * 3* 2 * 1) = (7 * 5)/(16 * 16) pi = (35pi)/(256) ` `[because sqrt(1/2) = sqrtpi]`

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