Special Integral Forms MCQ Quiz in मराठी - Objective Question with Answer for Special Integral Forms - मोफत PDF डाउनलोड करा

Concept:

Sin (A + B) = sin (A) cos (B) + cos (A) sin (B)

\(\smallint {\rm{cosec\;xdx}} = - \ln \left| {{\rm{cosec\;x}} + \cot {\rm{x}}} \right| + {\rm{c}} = \ln \left| {\tan \left( {\frac{{\rm{x}}}{2}} \right)} \right| + {\rm{c}}b\)

Calculation:

\({\rm{I}} = \smallint \frac{{{\rm{dx}}}}{{{\rm{acosx}} + {\rm{b}}\sin {\rm{x}}}}\)

Let, a = r sin α and b = r cos α

∴ \({\rm{I}} = \smallint \frac{{{\rm{dx}}}}{{{\rm{acosx}} + {\rm{b}}\sin {\rm{x}}}}\)

\( = \smallint \frac{{{\rm{dx}}}}{{{\rm{rsin\alpha cosx}} + {\rm{rcos\alpha }}\sin {\rm{x}}}}\)

\( = \smallint \frac{{{\rm{dx}}}}{{{\rm{r}}\left( {\sin \left( {{\rm{\alpha }} + {\rm{x}}} \right)} \right)}}\)

\( = \frac{1}{{\rm{r}}}\smallint {\rm{cosec\;}}\left( {{\rm{\alpha }} + {\rm{x}}} \right){\rm{dx}}\)

\( = \frac{1}{{\rm{r}}}\ln \left| {\tan \left( {\frac{{{\rm{\alpha }} + {\rm{x}}}}{2}} \right)} \right|\)

We have, a = r sin α and b = r cos α

Squaring and adding,

\({{\rm{a}}^2} + {{\rm{b}}^2} = {{\rm{r}}^2}\left( {{{\sin }^2}{\rm{\alpha }} + {{\cos }^2}{\rm{\alpha }}} \right)\)

\( \Rightarrow {\rm{r}} = \sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} \)

Hence, option (b) is correct.

Concept:

\(\smallint \frac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = {\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + C\)

Calculation:

Here we have to find the value of \(\smallint \frac{{dx}}{{\sqrt {9 - 25{x^2}} }}\)

The given integrand can be re-written as:

\(\Rightarrow \smallint \frac{{dx}}{{\sqrt {9 - 25{x^2}} }} = \frac{1}{5} \cdot \smallint \frac{{dx}}{{\sqrt {\frac{9}{{25}} - {x^2}} }}\)

\(= \frac{1}{5} \cdot \;\smallint \frac{{dx}}{{\sqrt {{{\left( {\frac{3}{5}} \right)}^2} - {x^2}} }}\)

Now by comparing,

 \(\smallint \frac{{dx}}{{\sqrt {{{\left( {\frac{3}{5}} \right)}^2} - {x^2}} }}\) with \(\smallint \frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}\) we get, a = 3/5

As we know that,

 \(\smallint \frac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = {\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + C\) where C is a constant.

\(\Rightarrow \smallint \frac{{dx}}{{\sqrt {9 - 25{x^2}} }} = \frac{1}{5} \cdot {\sin ^{ - 1}}\left( {\frac{{5x}}{3}} \right) + C\) where C is a constant

Concept:

• \(\rm \int\frac{dx}{\sqrt {x^{2}+a^{2}}}=log{\left |x+\sqrt {x^{2}+a^{2}}\right |}+C\)
• \(\rm \int_{0}^{x}{f(x) dx}=F(x)-F(0)\), where F(x) is the anti-derivative of f(x).

Calculation:

Given: \(\rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}+16}}\)

Using the formula, \(\rm \int\frac{dx}{\sqrt {x^{2}+a^{2}}}=log{\left |x+\sqrt {x^{2}+a^{2}}\right |}+C\)

\(\Rightarrow \rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}+16}}=\left [ log{\left |x+\sqrt {x^{2}+16}\right |} \right ]_{4}^{5}\)

​\(\Rightarrow \rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}+16}}=log{|5+\sqrt {5^{2}+16}}|-log{|4+\sqrt {4^{2}+16}}|\)​

\(=log{|5+\sqrt {41}}|-log{|4+\sqrt {32}}|\)

It is known that log a - log b = log (a / b)

\(\Rightarrow \rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}-16}}=log\frac{ {|5+\sqrt {41}}|}{ {|4+\sqrt {32}}|}\)

Hence, the correct answer is option 1.

Concept:

• \(\smallint \frac{{dx}}{{\sqrt {{x^2} - {a^2}} }} = \log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C\)

Calculation:

Here we have to find the value of \(\smallint \frac{{{x^2}}}{{\sqrt {{x^6} - 1} }}dx\)

Let x³ = t then 3x² dx = dt

\(\Rightarrow \smallint \frac{{{x^2}}}{{\sqrt {{x^6} - 1} }}dx = \frac{1}{3}\;\smallint \frac{{dt}}{{\sqrt {{t^2} - 1} }}\)

Now by comparing \(\;\smallint \frac{{dt}}{{\sqrt {{t^2} - 1} }}\) with \(\smallint \frac{{dx}}{{\sqrt {{x^2} - {a^2}} }}\) we get, a = 1

As we know that, \(\smallint \frac{{dx}}{{\sqrt {{x^2} - {a^2}} }} = \log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C\) where C is a constant

\(\Rightarrow \frac{1}{3}\;\smallint \frac{{dt}}{{\sqrt {{t^2} - 1} }} = \frac{1}{3}\log \left| {t + \sqrt {{t^2} - 1} } \right| + C\)

Now by substituting x³ = t in the above equation we get,

\(\Rightarrow \smallint \frac{{{x^2}}}{{\sqrt {{x^6} - 1} }}dx = \frac{1}{3}\log \left| {{x^3} + \sqrt {{x^6} - 1} } \right| + C\) where C is a constant

Concept:

• \(\smallint \frac{{dx}}{{{x^2} + {a^2}}} = \frac{1}{a}{\tan ^{ - 1}}\frac{x}{a} + C\)

Calculation:

Here we have to find the value of \(\smallint \frac{{dx}}{{4 + 25{x^2}}}\)

The given integrand can be rewritten as:

\(\Rightarrow \smallint \frac{{dx}}{{4 + 25{x^2}}} = \frac{1}{{25}}\;\smallint \frac{{dx}}{{\left( {\frac{4}{{25}} + {x^2}} \right)}}\)

\(= \frac{1}{{25}}\;\smallint \frac{{dx}}{{\left\{ {{{\left( {\frac{2}{5}} \right)}^2} + {x^2}} \right\}}}\)

By comparing \(\;\smallint \frac{{dx}}{{\left\{ {{{\left( {\frac{2}{5}} \right)}^2} + {x^2}} \right\}}}\) with \(\smallint \frac{{dx}}{{{x^2} + {a^2}}}\) we get a = 2/5

Now as we know that, \(\smallint \frac{{dx}}{{{x^2} + {a^2}}} = \frac{1}{a}{\tan ^{ - 1}}\frac{x}{a} + C\) where C is a constant

\(\Rightarrow \smallint \frac{{dx}}{{4 + 25{x^2}}} = \frac{1}{{10}}{\tan ^{ - 1}}\left( {\frac{{5x}}{2}} \right) + C\)

Concept:

\(\rm \int \frac{1}{x^{2}-a^{2}}dx = \frac{1}{2a}log\left | \frac{x-a}{x+a} \right |\) 

Calculation:

I = \(\rm \int \frac{dx}{x(x^6+1)}\)

Let x⁶ + 1 = t               ....(i)

Differentiate both sides w.r.t x , 

\(\rm \frac{\mathrm{d} \left ( x^{6}+1 \right )}{\mathrm{d} x}= \frac{\mathrm{d} t}{\mathrm{d} x}\)

⇒ 6 x⁵ dx = dt

⇒ dx = \(\rm \frac{dt}{6x^{5}}\)            

Putting above values in given integration, 

I = \(\rm \frac{1}{6}\int \frac{1}{tx^{6}}dt\) 

I = \(\rm \frac{1}{6}\int \frac{1}{t(t-1)}dt\) 

I = \(\rm \frac{1}{6}\int \frac{1}{t^{2}-t}dt\) 

By compleating square Method , 

I = \(\rm \frac{1}{6}\int \frac{1}{t^{2}-t+\frac{1}{4}-\frac{1}{4}}dt\) 

I = \(\rm \frac{1}{6}\int \frac{1}{(t-1/2)^{2}-(1/2)^{2}}dt\)  

I = \(\rm \frac{1}{6}log\left | \frac{t-1}{t} \right |+C\)

From eq . ( i ) 

⇒ I = \(\rm \frac{1}{6} \ln \left(\frac{x^6}{x^6 +1}\right)+C\) .

The correct option is 1.

Calculation:

We have, a = r sin α and

b = r cos α

\(\therefore \frac{{\rm{a}}}{{\rm{b}}} = \frac{{{\rm{rsin\alpha }}}}{{{\rm{rcos\alpha }}}}\)

\(\Rightarrow \frac{{\rm{a}}}{{\rm{b}}} = \tan {\rm{\alpha }}\)

\( \Rightarrow {\rm{\alpha }} = {\tan ^{ - 1}}\left( {\frac{{\rm{a}}}{{\rm{b}}}} \right)\)

Hence, option (1) is correct.

Concept:

• \(\rm \int\frac{dx}{\sqrt {x^{2}+a^{2}}}=log{\left |x+\sqrt {x^{2}+a^{2}}\right |}+C\)
• \(\rm \int_{0}^{x}{f(x) dx}=F(x)-F(0)\), where F(x) is the anti-derivative of f(x).

Calculation:

Given: \(\rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}+9}}=log\frac{ {|a+\sqrt {b}}|}{ {|c+\sqrt {d}}|}\)

Using the formula, \(\rm \int\frac{dx}{\sqrt {x^{2}+a^{2}}}=log{\left |x+\sqrt {x^{2}+a^{2}}\right |}+C\)

\(⇒ \rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}+9}}= \left [ log{\left |x+\sqrt {x^{2}+9}\right |} \right ]_{4}^{5}\)

​\(⇒ \rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}+9}}=log{|5+\sqrt {5^{2}+9}}|-log{|4+\sqrt {4^{2}+9}}|=log{|5+\sqrt {34}}|-log{|4+\sqrt {25}}|\)​

It is known that log a - log b = log a / b

\(⇒ \rm \int_{5}^{6} \frac{dx}{\sqrt {x^{2}-16}}=log\frac{ {|5+\sqrt {34}}|}{ {|4+\sqrt {25}}|}\)--------(1)

∵ It is given that, \(\rm \int_{4}^{5} \frac{dx}{\sqrt {x^{2}+9}}=log\frac{ {|a+\sqrt {b}}|}{ {|c+\sqrt {d}}|}\)----------(2)

On comparing (1) and (2) we get, a = 5, b = 34, c = 4, d = 25

⇒ a + b + c+ d = 5 + 34 + 4 + 25 = 68

Hence, the correct answer is option 2.

Concept:

• \(\rm \int\frac{dx}{\sqrt {x^{2}-a^{2}}}=log{\left |x+\sqrt {x^{2}-a^{2}}\right |}+C\)
• \(\rm \int_{0}^{x}{f(x) dx}=F(x)-F(0)\), where F(x) is the anti-derivative of f(x).

Calculation:

Given: \(\rm \int_{5}^{6} \frac{dx}{\sqrt {x^{2}-16}}\)

Using the formula, \(\rm \int\frac{dx}{\sqrt {x^{2}-a^{2}}}=log{\left |x+\sqrt {x^{2}-a^{2}}\right |}+C\)

\(\Rightarrow \rm \int_{5}^{6} \frac{dx}{\sqrt {x^{2}-16}}= \int_{5}^{6} \frac{dx}{\sqrt {x^{2}-4^{2}}}=\left [ log{\left |x+\sqrt {x^{2}-16}\right |} \right ]_{5}^{6}\)

Now, Substitute the limit to evaluate the value.

​\(\Rightarrow \rm \int_{5}^{6} \frac{dx}{\sqrt {x^{2}-16}}=log{|6+\sqrt {6^{2}-16}}|-log{|5+\sqrt {5^{2}-16}}|=log{|6+\sqrt {20}}|-log{|8}|\)​

It is known that log a - log b = log (a / b)

\(\Rightarrow \rm \int_{5}^{6} \frac{dx}{\sqrt {x^{2}-16}}=log\frac{ {|6+\sqrt {20}}|}{{8}}\)

Hence, the correct answer is option 4.

Concept:

• \(\rm \int \frac{dx}{\sqrt {a^{2}-x^{2}}}=sin^{-1}\frac{x}{a}+C\)
• \(\rm \int_{0}^{x}{f(x) dx}=F(x)-F(0)\), where F(x) is the anti-derivative of f(x).

Calculation:

Given: \(\rm \int_{0}^{1} \frac{dx}{\sqrt {1^{2}-x^{2}}}\)

Using the formula, \(\rm \int \frac{dx}{\sqrt {a^{2}-x^{2}}}=sin^{-1}\frac{x}{a}+C\)

\(\rm \Rightarrow \int \frac{dx}{\sqrt {1-x^{2}}}=\int \frac{dx}{\sqrt {1^{2}-x^{2}}}=\left [ sin^{-1}x \right ]_{0}^{1}=sin^{-1}1-sin^{-1}0\)

It is known that, \(\rm sin^{-1}0=0\)

\(\rm \Rightarrow \int \frac{dx}{\sqrt {1-x^{2}}}=sin^{-1}1\)

Hence, the correct answer is option 1.