Integrals- 3 - JEE Maths Free MCQ Test with solutions

I = ∫√(x² + 5x)dx

= ∫√(x² + 5x + 25/4 - 25/4)

= ∫√{(x + 5/2)² - (5/2)²}

={1/2(x+5/2)(√x² + 5x)} - {25/8 log{(x + 5/2)+√x²+ 5x}}

= {(2x + 5)/4 (√x² + 5x)} - {25/8 log{(x + 5/2)+√x²+ 5x}}

Thus, option D is correct...

sin²x = 1 - cos²x
∫sinx(sin²x - 3cos²x + 15)dx
Put cos²x = t
 ∫sinx(1 - cos²x - 3cos²x + 15)dx
=  ∫sinx (16 - 4cos²x)dx
Put t = cosx, differentiate with respect to x, we get 
dt/dx = -sinx
= -  ∫ [(16 - 4t²)]^1/2dt
= -2 ∫ [(2)² - (t)²]^½
= -2{[(2)² - (t)²]^½ + 2sin^-1(t/2)} + c
= - cosx {[4 - (cos)²x]^½ - 4sin^-1(cosx/2)} + c

Therefore , the answer is -cos(x³) + c

 (x)^½ (a - x)^½ dx
=  ∫(ax - x²)^½ dx
=  ∫{-(x² - ax)^½} dx
=  ∫{-(x² - ax + a^2/4 - a^2/4)^½} dx
=  ∫{-(x - a/2)² - a^2/4} dx
=  ∫{(a/2)² - (x - a/2)²} dx
=  ½(x - a/2) {(a/2)² - (x - a/2)²} + (a^2/4) (½ sin^-1(x - a)/a²)
= {(2x - a)/4 (ax - x²)^½} + {a^2/8 sin^-1(2x - a)/a} + c

None

Apply Trig Substitution: x=4sin(u)​

Let y = e^x
dy/dx = e^x dx
= ∫(9 - y²)^½ dy
= ∫[(3)² - (y)²]^½ dy
Apply [(a)² - (x)²] formula
⇒ y/2[(3)² - (y)²]^½ + [(3)²]/2 sin^-1(y/3) + c
= e^x/2[(3)² - (y)²]^½ + [9]/2 sin^-1(e^x/3) + c

= π / 4