Mathematics

# $\int sin^3 x cos^2x dx$=

$\dfrac{cos^5 x}{5} - \dfrac{cos^3x}{3} +c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Subjective Medium
$\displaystyle\int_{0}^{1} x^2-3x \ dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Evaluate: $\displaystyle \int \sqrt{x^{2}-x+1}dx.$
• A. $\displaystyle\frac{2x+1}{4}\sqrt{x^{2}-x+1}+\frac{3}{8}\log \left \{ \frac{2x+1}{2}+\sqrt{x^{2}-x+1} \right \}$
• B. $\displaystyle\frac{2x+1}{2}\sqrt{x^{2}-x+1}+\frac{3}{8}\log \left \{ \frac{2x-1}{4}+\sqrt{x^{2}-x+1} \right \}$
• C. $\displaystyle\frac{2x-1}{3}\sqrt{x^{2}-x+1}-\log \left \{ \frac{2x-1}{2}+\sqrt{x^{2}-x+1} \right \}$
• D. $\displaystyle\frac{2x-1}{3}\sqrt{x^{2}-x+1}+\frac{3}{8}\log \left \{ \frac{2x-1}{2}+\sqrt{x^{2}-x+1} \right \}$
• E. $\displaystyle\frac{2x-1}{2}\sqrt{x^{2}-x+1}+\frac{3}{8}\log \left \{ \frac{2x-1}{4}+\sqrt{x^{2}-x+1} \right \}$
• F. $\displaystyle\frac{2x-1}{4}\sqrt{x^{2}-x+1}-\frac{3}{8}\log \left \{ \frac{2x-1}{2}+\sqrt{x^{2}-x+1} \right \}$
• G. $\displaystyle\frac{2x-1}{4}\sqrt{x^{2}-x+1}+\frac{3}{8}\log \left \{ \frac{2x-1}{2}+\sqrt{x^{2}-x+1} \right \}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate: $\displaystyle {\int {\left( {\dfrac{{1 - x}}{{1 + {x^2}}}} \right)} ^2}{e^x}\,dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int _{ 0 }^{ a }{ \frac { dx }{ a+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } }$ is equal to
• A. $\displaystyle \frac { \pi }{ 2 } +1$
• B. $\displaystyle 1-\frac { \pi }{ 2 }$
• C. none of these
• D. $\displaystyle \frac { \pi }{ 2 } -1$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$