Solve the following integration ∫sin2x+sin5x−sin3xcosx+1−2sin22xdx View Solution Integrate:∫sinxcos2xsin3xdx View Solution Integrate:sinxsin(cosx) View Solution Integrate:sinxsin(cosx) Integrate:sinxsin(cosx) View Solution Doubtnut is No.1 Study App and Learning App with Instant Video...
1. Trigonometric identity: cos2(x)=1+cos(2x)2.2. Move the constant out: ∫b⋅f(x)dx=b⋅∫f(x)dx.3. Common integration: ∫cos(u)du=sin(u).4. The sum rule: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx....
Step 2: Use the integration formula for sineWe know that the integral of sin(kx) is given by:∫sin(kx)dx=−1kcos(kx)+Cwhere k is a constant and C is the constant of integration. Step 3: Apply the formulaIn our case, k=2. Therefore, we apply the formula:I=−12cos(2x)+C...
sin2x的不定积分公式：∫xsin2xdx=（-1/2）∫xdcos2x。在微积分中，一个函数f的不定积分，或原函数，或反导数，是一个导数等于f的函数F，即F′=f。不定积分和定积分间的关系由微积分基本定理确定。其中F是f的不定积分。微积分（Calculus），数学概念，是高等数学中研究函数的微分（Differentiati...
Evaluate: ∫sin2xdx. Integration Of A Function: Integration is the inverse process of differentiation. Consider a function g such that g(x)=G′(x), then ∫g(x)dx=G(x)+C. Useful Formula: cos2x=1−2sin2xsin2x=1−cos2x2. Answer and Explanation: Let I=∫...
Use integration by parts to find the integral. (Use C for the constant of integration.) {eq}\displaystyle \int x \sin^2 (x)\ dx {/eq}. Integration by parts: If {eq}f(x) {/eq} and {eq}g(x) {/eq} are two functions, then {eq}\int f...
∫xsin2xdx=（-1/2）∫xdcos2x。在微积分中，一个函数f的不定积分，或原函数，或反导数，是一个导数等于f的函数F，即F′=f。不定积分和定积分间的关系由微积分基本定理确定。其中F是f的不定积分。微积分（Calculus），数学概念，是高等数学中研究函数的微分（Differentiation）、积分（Integration...
Using binomial expansion, we have that (1+x)k=∑kj=0(jk)xj(1+x)k=∑j=0k(jk)xj. To get the 12j+112j+1 factor, we have to do some integration so with this in mind, we have that (1−x2)k=∑j=0k(jk)(−1)jx2j(1−x2)k=∑j=0k(jk)(−...
whereCis the constant of integration. 5.Combine Results: Substitute back into the equation: 6.Final Result: The final result of the integral is: Thus, the final result is: −cos2(x)2 Integral Calculatorcomputes an indefinite integral (anti-derivative) of a function with respect to a given...
I=2∫12√0arcsinx12−x2−−−−−−√dx=2Li2(12–√)−π224+ln224I=2∫012arcsinx12−x2dx=2Li2(12)−π224+ln224 Where the latter integral was evaluated with wolfram. I would love to see a proof for that. integration power-series closed-form Share ...