Mathematics

# Evaluate: $\int \sqrt{1+sinx}dx$

##### SOLUTION
$\int \sqrt{1+sin x}dx$      Replace $1\rightarrow sin^{2}x cos^{2}x$
$= \int \sqrt{sin^{2}x+cos^{2}x+2 sin\frac{x}{2}}dx$    (using submultiple angle formula)
$=\int \sqrt{(sin x+cos x)^{2}}dx$
$=\int (sin x+cos x)dx$
$=-cos x+sin x+c$
Where c is the constant of integration.

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

#### Realted Questions

Q1 Single Correct Hard
If $\displaystyle \frac{2x}{x^3-1}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$, then
• A. $A=B\neq C$
• B. $A\neq B=C$
• C. $A\neq B \neq C$
• D. $A=-B=C$

1 Verified Answer | Published on 17th 09, 2020

Q2 One Word Hard
The value of the integral
$\displaystyle \int_{0}^{\frac {1}{2}}\frac {1 + \sqrt {3}}{(x + 1)^{2} (1 - x)^{6})^{\frac {1}{4}}}dx$ is _______.

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate $\displaystyle \int_{0}^{2}(x^2+1)dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $\displaystyle \int \dfrac{2^x}{\sqrt{1 - 4^x}} dx = \lambda \sin^{-1} 2^x + C$, then $\lambda$ equals to
• A. $\log 2$
• B. $\dfrac{1}{2} \log 2$
• C. $\dfrac{1}{2}$
• D. $\dfrac{1}{\log 2}$

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 $\displaystyle f(x)>0 \forall x \in R$