Evaluate using Trigonometric Identities MCQ Quiz - Objective Question with Answer for Evaluate using Trigonometric Identities - Download Free PDF

\(I=\displaystyle\int_0^\pi\sqrt{1+4\sin^2(x/2)-4\sin(x/2)}dx\)

\(I=\displaystyle\int_0^\pi|2\sin(x/2)-1|dx\)

If x = 0, then \(2\sin (x/2)-1<0\)

And if x = \(\dfrac{\pi}{3}\), then \(2\sin (x/2)-1>0\)

\(\therefore I=\displaystyle\int_0^{\pi/3}-(2\sin(x/2)-1)dx+\displaystyle\int_{\pi/3}^{\pi}(2\sin(x/2)-1)dx\)

\(I=[4cos(x/2)+x]_0^{\pi/3}+[-4\cos (x/2)-x]_{\pi/3}^\pi\)

\(I=\dfrac{4\sqrt3}{2}+\dfrac{\pi}{3}-4+\Big(0-\pi+\dfrac{4\sqrt3}{2}+\dfrac{\pi}{3}\Big)\)

\(I=4\sqrt3-4-\dfrac{\pi}{3}\)

Concept:

The given integral is: \( \int_{-1}^{1} \int_{0}^{\sqrt{1 - x^2}} (x^2 + y^2)^{\frac{3}{2}} \, dy\, dx \)

This represents a double integral over the upper half of the unit circle centered at the origin. So, converting it into polar coordinates simplifies the integration process.

Let \( x = r\cos\theta,\ y = r\sin\theta \) and \( dx\,dy = r\,dr\,d\theta \)

The region of integration becomes a semicircle: \( r \in [0,1],\ \theta \in [0,\pi] \)

Calculation:

In polar coordinates, \( x^2 + y^2 = r^2 \Rightarrow (x^2 + y^2)^{3/2} = r^3 \)

So, the integral becomes:

\( \int_{0}^{\pi} \int_{0}^{1} r^3 \cdot r \, dr\, d\theta = \int_{0}^{\pi} \int_{0}^{1} r^4 \, dr\, d\theta \)

\(= \int_{0}^{\pi} \left[\frac{r^5}{5}\right]_{0}^{1} d\theta = \int_{0}^{\pi} \frac{1}{5} \, d\theta = \frac{\pi}{5} \)

Correct Option:

The correct answer is: 3) π/5

Explanation:

We are given the integral:

Integral: I = \(\int_{0}^{1} \frac{a-b x^{2}}{\left(a+b x^{2}\right)^{2}} d x\) 

Step 1: Substitution

Let u = a + bx², then du = 2bx dx.

Hence, x dx = du / (2b).

After substitution, the limits change: when x = 0, u = a, and

when x = 1, u = a + b.

The integral becomes:

I = 1 / (2b) ∫_a^a+b (2a - u) / u² du

Step 2: Decompose the Integral

I = 1 / (2b) [ 2a ∫_a^a+b 1/u² du - ∫_a^a+b 1/u du ]

Each part is solved as follows:

• ∫_a^a+b 1/u² du = [ -1/u ]_a^a+b = 1/a - 1/(a+b)
• ∫_a^a+b 1/u du = ln(u)_a^a+b = ln(a+b) - ln(a)

Thus, the integral becomes:

I = 1 / (2b) [ 2a (1/a - 1/(a+b)) - (ln(a+b) - ln(a)) ]

Final Result:

The value of the integral is: I = 1 / (a + b)

Calculation

\(\mathrm{I}(\mathrm{~m}, \mathrm{~m})=\int_{0}^{1} \mathrm{x}^{\mathrm{m}-1}(1-\mathrm{x})^{\mathrm{n}-1} \mathrm{dx}\)

Let x = sin²θ dx = 2sinθcosθdθ

⇒ \(I(m, n)=2 \int_{0}^{\pi / 2}(\sin \theta)^{2 m-1}(\cos \theta)^{2 n-1} d \theta\)

⇒ I (9, 14) + I (10, 13) = \(2 \int_{0}^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{27} d \theta\)

\(+2 \int_{0}^{\pi / 2}(\sin \theta)^{19}(\cos \theta)^{25} d \theta\)

⇒ \(2 \int_{0}^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{25} d \theta\)

⇒ I (9, 13) 

Hence option 4 is correct

Calculation

\(\int x^{3} \sin x d x=-x^{3} \cos x+\int 3 x^{2} \cos x d x\)

= \(-x^{3} \cos x+3 x^{2} \sin x-\int 6 x \sin x d x\)

= -x³ cos x + 3x² sin x + 6x cos x - 6 sin x + c

So g(x) = -x3 cos x + 3x2 sin x + 6x cos x - 6 sin x 

\(g\left(\frac{\pi}{2}\right)=\frac{3 \pi^{2}}{4}-6\)

g'(x) = 3x² cos x + x³ sin x + 6cos x - 6cos x

\(\mathrm{g}^{\prime}\left(\frac{\pi}{2}\right)=\frac{\pi^{3}}{8}\)

\(8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\pi^{3}+6 \pi^{2}-48\)

So α + β – γ = 55

Hence option 1 is correct

Concept:

• \({\sin ^2}{\rm{x}} + {\cos ^2}{\rm{x}} = 1\)
• \(\sin 2{\rm{x}} = 2\sin {\rm{x}}\cos {\rm{x}}\)

Calculation:

Let, \({\rm{I}} = \mathop \smallint \limits_0^{1} \sqrt {1 + \sin \frac{{\rm{x}}}{2}} {\rm{dx}}\)

\(= \mathop \smallint \limits_0^1 \sqrt {{{\sin }^2}\left( {\frac{{\rm{x}}}{4}} \right) + {{\cos }^2}\left( {\frac{{\rm{x}}}{4}} \right) + \sin \frac{{\rm{x}}}{2}} {\rm{dx}} \)        ----( \({\sin ^2}\left( {\frac{{\rm{x}}}{4}} \right) + {\cos ^2}\left( {\frac{{\rm{x}}}{4}} \right)\) = 1)

\(= \mathop \smallint \limits_0^1 \sqrt {{{\sin }^2}\left( {\frac{{\rm{x}}}{4}} \right) + {{\cos }^2}\left( {\frac{{\rm{x}}}{4}} \right) + 2\sin \left( {\frac{{\rm{x}}}{4}} \right)\cos \left( {\frac{{\rm{x}}}{4}} \right)} {\rm{dx}} \)       -----\(\left( {\sin \frac{{\rm{x}}}{2} = 2\sin \left( {\frac{{\rm{x}}}{4}} \right)\cos \left( {\frac{{\rm{x}}}{4}} \right)} \right)\)

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^1 \sqrt {{{\left( {\sin \left( {\frac{{\rm{x}}}{4}} \right) + \cos \left( {\frac{{\rm{x}}}{4}} \right)} \right)}^2}} {\rm{dx}}\)

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^1\left| {\sin \left( {\frac{{\rm{x}}}{4}} \right) + \cos \left( {\frac{{\rm{x}}}{4}} \right)} \right|{\rm{dx}}\)

Let \(\frac{{\rm{x}}}{4} = {\rm{t}}\)

Differentiating with respect to x, we get

\(\Rightarrow \frac{{{\rm{dx}}}}{4} = {\rm{dt}}\)                      

\(\Rightarrow {\rm{dx}} = 4{\rm{dt}}\)

Limits become for x = 0, t = 0

for x = 1, t = 1/4 = 0.25

\(\therefore {\rm{I}} = 4\mathop \smallint \limits_0^{0.25} \left| {\sin \left( {\rm{t}} \right) + \cos \left( {\rm{t}} \right)} \right|{\rm{dt}}\)

\(\Rightarrow {\rm{I}} = 4\left[ { - \cos {\rm{t}} + \sin {\rm{t}}} \right]_0^{0.25} \)

\(= 4\left[ {\left( { - \cos \left( {\frac{{\rm{1 }}}{4}} \right) + \sin \left( {\frac{{\rm{1 }}}{4}} \right)} \right) - \left( { - \cos 0 + \sin 0} \right)} \right] \)

\(= 4\left[ { 1 - \cos \left( {\frac{{\rm{1 }}}{4}} \right) + \sin \left( {\frac{{\rm{1 }}}{4}} \right)} \right] \)

Hence, option (4) is correct.

Concept:

• sin² x + cos² x = 1
• sin 2A = 2 sin A cos A
• ∫ cos x = sin x

• ∫ sin x = -cos x

Calculation:

\(∫_{0}^{\pi /3}\sqrt{1+\sin 2x}\, dx\)

\(\Rightarrow ∫_{0}^{\pi /3}\sqrt{(\sin ^2x\, +\, \cos ^2x\, +\, 2\sin x\, \cos x})\, dx\)

\(\Rightarrow ∫_{0}^{\pi /3}(\cos x\, +\,\sin x)\, dx\)

\(\Rightarrow \left [ \sin x \, -\, \cos x \right ]_{0}^{\pi /3}\)

\(\Rightarrow \left [ \sin \frac{\pi }{3}\, -\, \cos \frac{\pi }{3} \right ]-(\sin 0\, -\, \cos 0) \)

\(\Rightarrow \left [ \frac{\sqrt{3} }{2}\, -\, \frac{1 }{2} \right ]-(0\, -\, 1)\)

\(\Rightarrow \left [ \frac{\sqrt{3}-1 }{2} \right ]+ 1\)

\(\Rightarrow \frac{\sqrt{3}+1}{2}\)

Hence, \(∫_{0}^{\pi /3}\sqrt{1+\sin 2x}\, dx=\frac{\sqrt{3}+1}{2}\)

Concept:

1 + tan² x = sec² x

\(\rm \displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+c\)

Calculation:

\(\text{Let I} = \rm \displaystyle \int_0^{π/4} (\tan^3 x + \tan x)dx\\\Rightarrow\rm \displaystyle \int_0^{π/4} \tan x(\tan^2 x + 1)dx\)

\(\\\Rightarrow \rm \displaystyle \int_0^{π/4} \tan x \sec^2 x dx\)             (∵ 1 + tan² x = sec² x)

Let tanx = t

Differentiating with respect to x, we get

⇒ sec2 x dx = dt

\(\Rightarrow\rm \displaystyle \int_0^1 tdt\\\Rightarrow[\frac{t^2}{2}]_0^1\\\Rightarrow\frac{1}{2}-0=\frac{1}{2}\)

∴ The value of the integral \(\rm \displaystyle\int_0^{\pi/4}(\tan^3 x + \tan x)dx \ \)is 1/2.

Concept:

• \({\sin ^2}{\rm{x}} + {\cos ^2}{\rm{x}} = 1\)
• \(\rm \cos 2{\rm{x}} = \cos^2 x - \sin^2x\)

Calculation:

 Given: \({\rm{I}} = \mathop \smallint \limits_0^{2{\rm{\pi }}} √ {1 + \cos 2x} {\rm{dx}}\)

\(= \mathop \smallint \limits_0^{2{\rm{\pi }}} √ {{{\sin }^2}\ x + {{\cos }^2}\ x + \cos^2 x - \sin^2x}\)                  (1 = \({\sin ^2}x + {\cos ^2}x\))

\(= \mathop \smallint \limits_0^{2{\rm{\pi }}} √ {2 \cos^2 x}\ dx\)             

= \(\rm √ 2 \mathop \smallint \limits_0^{2{\rm{\pi }}} \cos x \ dx\)

= \(\rm √ 2 \left[ {\sin x} \right]_0^{{{\rm{2\pi }}}}\)

= \(\rm √ 2 [\sin {2\pi\ -\ \sin 0}]\)

= 0

Concept:

Definite Integral:

If ∫ f(x) dx = g(x) + C, then \(\rm \int_a^b f(x)\ dx = [ g(x)]_a^b\) = g(b) - g(a).

Trigonometric Identities:

cos 2x = 2 cos² x - 1

Calculation:

Let I = \(\rm \int_{0}^{\pi/2}\cos^2x\ dx\)

⇒ I = \(\rm \int_{0}^{\pi/2}\frac{\cos 2x+1}{2}\ dx\)

⇒ I = \(\rm \frac{1}{2}\left[\int_{0}^{\pi/2}\cos 2x\ dx+\int_{0}^{\pi/2}1\ dx\right]\)

⇒ I = \(\rm \frac{1}{2}\left[\frac{\sin 2x}{2}+x\right]_0^{\pi/2}\)

⇒ I = \(\rm \frac{1}{2}\left[\left(\frac{\sin \pi}{2}+\frac{\pi}{2}\right)-\left(\frac{\sin 0}{2}+0\right)\right]\)

⇒ I = \(\frac\pi4\).

Concept:

\(\rm {\sin ^2}{\rm{x}} + {\cos ^2}{\rm{x}} = 1\)

\(\rm\cos 2{\rm{x}} = cos^2{x} - sin^2{x}\)

Calculation:

Let, \({\rm{I}} = \mathop \smallint \limits_0^{2{\rm{\pi }}} \sqrt {1 + \cos \frac{{\rm{x}}}{2}} {\rm{dx}}\)

\(⇒ {\rm{I}} = \mathop \smallint \limits_0^{2{\rm{\pi }}} \sqrt {(\rm 2cos^2{(\frac {x}{4})}})\)

\(⇒ {\rm{I}} =\sqrt 2 \mathop \smallint \limits_0^{2{\rm{\pi }}} \cos{(\frac {\rm x}{4})}\)

\(⇒ {\rm{I}} = 4\sqrt 2\left[ { \sin {\rm{(\frac {x}{4}})}} \right]_0^{2\pi}\)

\(⇒ {\rm{I}} = 4\sqrt 2[ { \sin {\rm{(\frac {2\pi}{4}}) - sin0^o}}]\)

\(⇒ {\rm{I}} = 4\sqrt 2 (\sin {\frac {\pi}{2}})\)

\(⇒ {\rm{I}} = 4 \sqrt 2\)

Hence, option (4) is correct.

Concept:

sin² x + cos² x = 1

\(\rm \frac{d(sinx)}{dx} = cos x\)

\(\rm \int x^ndx = \frac{x^{n + 1}}{n + 1}\)

\(\rm \int dx = x\)

Calculation:

I = \(\rm \int_{0}^{\pi/2 } cos^3 x dx\)

= \(\rm \int_{0}^{\pi/2 } cos^2 x .cos x dx\)

= \(\rm \int_{0}^{\pi/2 } (1 - sin^2 x) .cos x dx\)

Let, sin x = u

cos x dx = du

= \(\rm \int_{0}^{1} (1 - u^2)du\)

= \(\rm \left [u - \frac{u^{3}}{3} \right ]_{0}^{1}\)

= 1 - 0 - \(\rm \frac{1}{3} + 0\)

= \(\rm \frac{2}{3}\)

Formula used:

\(\rm \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx\)

tan(π - θ) = - tan θ  

Calculation:

Let

I = \(\int^π _0 ln\left(tan\frac{x}{2}\right)dx\)        ---(1)

According to the formula used

I = \(\int^π _0 ln\left(tan (π -\frac{x}{2})\right)dx\)

⇒ I = - \(\int^π _0 ln\left(tan\frac{x}{2}\right)dx\)

From equation (1)

⇒ I = -I

⇒ 2I = 0

⇒ I = 0

∴ The value of the integral \(\int^π _0 ln\left(tan\frac{x}{2}\right)dx\) is 0.

Concept:

Definite Integral:

If ∫ f(x) dx = g(x) + C, then \(\rm \int_a^b f(x)\ dx = [ g(x)]_a^b\) = g(b) - g(a).

Calculation:

Let I = \(\rm \int_{0}^{\pi/2}\cos 2x\ dx\)

⇒ I = \(\rm \left[\frac{\sin 2x}{2}\right]_0^{\pi/2}\)

⇒ I = \(\rm \left[\frac{\sin \pi}{2}-\frac{\sin 0}{2}\right]\)

⇒ I = 0.

Concept:

cot² x = cosec² x - 1 

\(\rm \int x^{n}dx = \frac{x^{n+1}}{n+1} + C\) 

Calculation:

I = \(\rm \int_0^{π/4}(\cot^3 x + \cot x)dx \ \) 

I = \(\rm\int_{0}^{π /4}\left [ \left ( cosec^{2}x-1 \right )cotx+ cotx \right ]dx\) 

I = \(\rm \int_{0}^{π /4}\left [ cosec^{2}x. cotx-cotx+cotx \right ]dx\) 

I = \(\rm \int_{0}^{π /4}\left ( cosec^{2}x.cotx \right )dx\) 

I = \(\rm \int_{0}^{π /4}\left ( \frac{1}{sin^{2}x} . \frac{cosx}{sinx}\right )dx\) 

I = \(\rm\int_{0}^{π/4}\left ( \frac{cosx}{sin^{3}x} \right )dx\)           ...(1)

Let , sin x = t 

Differentiate both side  w.r.t  x , 

cos x dx = dt   

If , x = 0 , then t = 0 

x = π /4 , then t = \(\frac{1}{\sqrt{2}}\) 

substitute above values in eq. (1) 

I = \(\rm\int_{0}^{\frac{1}{\sqrt{2}}}\left ( \frac{dt}{t^{3}} \right )\)  

I = \(\rm \left [ \frac{1}{-2t^{2}} \right ]_{0}^{\frac{1}{\sqrt{2}}}\) 

I = \(\rm \frac{-1}{2}\left [ 2 -∞ \right ]\)  

I = ∞  

∴ The correct option is 4).